all right so uh let's start uh the problem solving session for chapter 14 okay so let's start with uh question number one okay so what we are given here and what it asks us to calculate let's see okay so how much heat input is needed okay so here it asks us to determine how much heat okay so that's the unknown and what are the given thing here so it says that to raise the temperature of um of an empty 20 kilogram fat made of iron from 19 degree a i mean 10 degrees celsius to 90 degree celsius okay so the first one is when the vat is empty and the second one is if the vat is filled with 20 kilogram of water okay in both cases we have to calculate the the heat required that means we have to calculate q okay so first let's do the first one so q vat okay now we know that m the definition is m c delta t right uh for uh here m is the uh the mass of the vat that is 20 kilogram okay and see it's made of iron right now we need to know what is the iron's specific heat capacity and i already have it you can check the table in the lecture and that is equal to 450 joule per kilogram per degree celsius and then delta t is the final temperature minus initial temperature right final temperature is 90 degree celsius minus 10 degree celsius okay so if you multiply all these three number okay i mean if you do this math uh you should get uh you should get uh so i got 7.2 [Music] times 10 to the power 5 joule 7.2 times 10 to the power 5 joule okay so that is the that is for the first one and now second one okay so second one is basically this one the one we have already for the bat and then we have to add uh and add the heat required for 20 20 20 kilogram of water okay so let's first calculate separately for water and then we can add them together so for water it will be m c water delta t okay m is again 20 kilogram because we have 20 kilogram of water and for water cw that means the specific heat capacity is 4 1 86 joule kilogram degree celsius and this is again 90 degree celsius minus 10 degree celsius okay so this uh is equal to 6.7 times 10 to the power 6 joule okay so the total heat we need is q vat plus q water okay so when there is water in the vat the total heat we need is seven point two times ten to the power five joule plus six point seven times ten to the power six joule all together it is seven point four times 10 to the power 6 joule okay and this is the answer for the second one all right okay so now let's do the next one okay so here we are given uh so if 200 centimeter cube of t at 95 degrees celsius is poured into a 150 gram glass cup okay initially at 25 degree celsius okay so we have many thing already given so this is the volume of t actually uh so um this is the volume okay so this is the volume of t and the initial temperature of the t is 95 degree celsius okay and then this is the the mass of the cup and the initial temperature of the cup okay now here you see that when you put uh the hot cup uh i mean hot tea on the car so the t is at 95 degree celsius that means it is at high temperature and cup is at uh lower temperature 25 degree celsius when the tea is on the cup you know that the tea will be uh losing uh heat okay it will be giving heat to the cup and cup will be gaining heat okay and because of the thermal equilibrium at some point the temperature of the both object i mean the t and the cup will be the same okay and that temperature we have calculate so what is the final temperature of the system of these two system okay right now there are many ways to calculate it so we'll be using uh the conservation of energy equation so you remember that conservation of energy is this one q equal to zero okay sum of q equal to zero okay so here we have uh two terms one is by a t another is by cop okay so one is by t another is by cup okay so q t plus q cup okay that is equal to zero now q t will be m t c water actually t is um almost water okay so we assume uh it is the same as water so the specific heat capacity of t is basically the specific heat of water okay and then delta t and the same similar thing for the cup m cup okay and glass so c glass and then delta t okay so this is the delta t for t and this is the delta t for cop okay now before we move uh to this equation first uh first thing is here the mass of the t is not given okay instead the volume of the t is given okay so we need the mass of the t so how to how to determine the mass from the volume okay so the equation we need here is this one okay the row the density of water is equal to mass per unit volume so mass is equal to density times the volume so as we are given the volume and we know the density of water already so we can multiply these two terms to get the uh to get the mass of the t so the mass of the t will be uh 1000 kilogram [Music] per meter cube so this is the water's density and v is the 200 centimeter cube okay now centimeter cube you know that's not the um not it's not the standard international unit okay so we have to convert centimeter cube to meter cube right so uh what we need is 10 to the power minus 2 3 times okay because we have centimeter cube okay so centimeter to meter uh is uh i mean one one centimeter is equal to 10 to the power minus 2 meter okay so using this you can show that you need basically three times this and that that is meter cube so what we have here if you do this math you get the mass of 200 meter centimeter cube t is equal to 0.20 kilogram okay so now um we know everything right so mt is now this kilogram specific heat of water is 4186 joule kilogram degree celsius okay delta t for t so final temperature for t we don't know okay so that we have to calculate initial temperature for the t is 95 degree celsius okay so you see that this is different than the these the change in temperature for the t is different the different than that change in temperature for the cup so for the cup it is um this kilogram right the value is given in grams we have to convert it into kilogram and for glass the value we can take from the table given in the uh given in the textbook and delta t here is a final is t we don't know and initially is 25 degree celsius okay so you will see that the fast term is positive first term is negative because the final temperature is lower than 95 degree celsius and that tells us that the t is leaving the heat is living from the t and the second term is positive and that tells us that it is entering the cup okay so we'll be using these equations and solve it for t this equation for t okay now um so we'll have four terms okay um so let me write down all the four terms and then do so minus so plus point one five then eight four zero t and then minus 0.15 840 and 25 equal to zero okay now you see that two terms uh we have two terms that contains t okay and then two other terms are completely numerical so if you simplify these you should get something like this so i got something like this 9 63 point 2 t minus 8 2 6 4 actually 6 8 4 and that is equal to 0 so t is equal to 8 2 6 8 4 divided by 963.2 okay and that is equal to 86 degrees celsius okay all right so there are some other way to you can solve these okay so in the textbook you will see that there are two methods they show but i prefer this one okay but yeah i mean no matter what method you use if you get the correct answer you get the full point okay all right the next one so here um an engineer which is to determine the specific heat so here we basically use the calorimetry okay so you want to know what is the specific heat of a sample okay so that we have to calculate and what we are given we are given uh the samples mass okay the samples initial temperature okay and then it says that this sample is quickly placed in point four kilogram of water and water's initial temperature is also given which is contained in a in a aluminum cup whose mass is 0.2 kilogram okay okay so that means um this the aluminium cups initial temperature is also 10 degree celsius okay and the final temperature of all the materials i mean the sample water and the aluminum cup is 30.5 degree celsius so if so what is the specific heat capacity of that sample okay now here you see that the sample is very high is initial the sample is at very high temperature right now when you put the sample in the water right the sample will lose heat okay on the other hand the water and and the aluminum cup will gain heat okay now here we will have three terms so in the last problem we basically solve a similar problem but um but there we had only two terms here we will have three terms okay one for the uh sample one for the water and one for the aluminum cup okay other than that it's exactly the same thing okay okay so so again let's start from this equation q equal to zero and then we'll have q sample okay and then q water and then q aluminium that is equal to zero right the three terms now for example it will be m sample c sample c sample you want to know what it is and then delta t sample for water m water sea water delta t for water the same thing for the cup aluminum cup right so it is m aluminium c aluminium delta t aluminum that is equal to zero okay okay so uh we have to we have to basically calculate cs so let me solve it for cs and then plug the value okay so it is mscs delta t s and that is equal to negative ah yeah negative m w c w delta t w negative m a c a delta t a okay so c s is equal to negative m w c w delta t w negative m a c a delta t a and then here m s delta t yes okay now we have to just plug the values okay so let's do it one by one mw means the water okay so water's mass is this water specific heat capacity is this and delta t for water is um water uh initially was at 10 degree celsius and final temperature is 30.5 degree celsius 30.5 minus 10 um 10 degrees celsius that's for water okay and this is also uh for the aluminum cup okay so the aluminum cups mass is this one okay aluminium's uh specific heat is 900 joule per kilogram degree celsius and delta t is the same thing for aluminum 30.5 degree celsius minus 10 degrees celsius okay so that's the numerator denominator is uh the sample's mass is 0.150 kilogram and final temperature is this initial temperature is 540 degree celsius okay now if you do the math in your calculator correctly you will see the value of cs i mean the specific heat capacity of the of the sample is this 497 joule per kilogram degree celsius okay all right so here you see that this term is negative and this negative will cancel the negative um in the numerator okay and that's why we get a positive value so make sure you understand how i calculate it okay all right let's move on all right so it's very similar this problem is also similar to the last one here we have how much energy does a freezer have to remove from one kilogram of water at 20 degree celsius to make it ice at 12 degree celsius so we have water at 20 degree celsius you know that if you want to decrease the temperature okay you need to remove the heat okay and the refrigerator can remove the heat from the sample okay and how much heat we have to remove that we have to calculate okay all right so um the energy need to remove here is equal to um so the heat required to remove the temperature i mean the the total heat we need is basically equal to the two heat one heat q1 and q2 let me talk later actually three heat okay and let me talk one by one so q1 is the heat that uh this uh this water i mean 1.5 kilogram water are um uh i mean to decrease the temperature of the 1.5 kilogram of water from 20 degree celsius to 0 degree celsius so this is the same mass for all okay so m will be the same for all the q one q two and q three and then c will be for waters specific heat and delta t will be uh the final uh final is zero degree celsius and initial is 20 degree celsius okay so this is the first term and this term tells us that uh the heat required to remove the temperature from 20 degrees celsius to zero degrees celsius so zero is final and 2 is initial that's why 0 minus 20 okay make sure the order um so the final minus the initial okay and then q2 is the mass times the latent heat of fusion okay so you know that at zero degree celsius uh in this case it will be freezing okay so in the theory class we see that um that it was melting at zero degree celsius because we are adding heat because here we are removing heat that means at zero degree celsius uh the the freezing is happening okay so we need these this term and then again we need another term to to remove uh i mean to decrease the temperature of the eyes so this water is now ice okay and we need we need some heat to remove from the eyes uh to decrease the temperature from zero degree celsius to minus 12 degree celsius okay so the first term is the for the is for the water second term is the is for the phase change and third term is for the eyes okay now we have to use the specific heat for the eyes and the final temperature is minus 12 degree celsius and initial temperature for this one for ice is zero degree celsius so again make sure you understand the order okay so here we have three terms one for the water another for the uh phase change and another for the eyes okay now if you plug all this value here now what we get 1.5 kilogram okay cw is 4.186 joule kilogram degree celsius and this is minus 20 degree celsius and then mass is again 1.5 kilogram the latent heat of fusion is 333 times 10 to the power 3 joule per kilogram okay make sure you understand this okay and then again this is 1.5 kilogram sea ice is 2100 joule kilogram degree celsius and this is minus 12 degree celsius okay now when you add all these three terms you should get something like this negative 6.6 times 10 to the power 5 joule okay so that tells us that heat is removing so i mean heat is leaving the system so minus sign tells us about that okay so negative means heat is leaving from the system okay all right let's move on um so another similar problem i mean and not a similar problem it is similar to um to uh two and three okay so here what it said it says that a at a reception okay um if 0.5 kilogram chunk of ice at negative 10 degree celsius is placed in three kilogram of iced tea okay so ice here and here we have water and that is at 20 degree celsius so we have two object one is ice there is water okay now at what temperature and in what phase will the final mixture uh mixture be okay so we we have to basically calculate the final temperature okay final temperature and in what phase it will be okay here you can consider the t as water and ignore any heat flow to the surrounding okay including the container okay so we can use the uh this equation q equal to zero okay now here when you um put the ice in the t okay so the t will um t will lose energy because it is at higher temperature than the ice and ice will gain energy right so q ice plus q water okay now q i's will have several term right y q i's will have several time term because ice is at 10 degrees minus 10 degree celsius so we'll have one term to raise the temperature from minus 10 degree celsius to zero degree celsius okay and then at zero degree celsius there will be phase change and then um and then from zero degree to some final temperature okay so for ice there will be three terms and for water there will be one term okay so for ice that will be one two three similar to the last problem okay so in the last problem uh heat removes from the system and here heat will adding to the ice from the water okay from the t actually okay so okay so first term so this this the parenthesis is for the eyes okay so first term is these okay and then i says um i says uh specific heat is 2100 kilogram sorry joule kilogram degree celsius and then for the eyes the final temperature is 0 degree celsius minus minus 10 degree celsius is the initial temperature okay so this is the first term and then the phase change will happen and then phase change means these times um okay this kilogram and then 333 times 10 to the power 3 joule per kilogram okay and then the last term for the eyes is again this kilogram and this time the the ice is basically water so we have to use this term okay and here the final temperature is we don't know the initial temperature is zero degree celsius so these three terms are for the eyes and for water or eyes we'll have 3 kilogram and then 4186 joule kilogram degree celsius final temperature we don't know initial temperature is 20 degree celsius okay so we have many terms here okay so let me write down so the first term will be something like this 0 5 2 1 0 0 and this is basically minus and minus is plus so it is 10 so this is simple the second term is also simple okay third term is also simple okay that is just t and fourth term is basically two terms okay so now we have to write two terms for water one for t another for 20 degree celsius so 3 4 1 8 6 t and minus 3 4 1 8 6 20 that is equal to zero okay so now we have five terms two terms contain t and all the other three terms are numerical terms so if you plug all these values in your calculator correctly you should get something like this actually i did all together so what i got final value t is equal to 5.1 degree celsius okay and this tells us that the final phase is the liquid phase okay okay final phase is liquid because uh at five degree celsius water is water basically okay if it is less than zero degree celsius then it is ice if it is higher than 100 degree celsius it will be steam okay in between 0 and 100 it is water okay so that means it is in the liquid phase okay all right all right so this is a conduction uh the rate of heat transfer by conduction method okay what we are given so we have to calculate the rate of heat flow okay rate of heat flow that means q over t we have to calculate and we are given we are given uh the area cross-sectional area thickness okay and the temperature between the two ends okay and this is a glass and for glass ah we know the thermal conductivity from the uh from the table okay so here is the equation q over t equal to k a t t1 minus t2 so t1 minus t2 is the final i mean the higher temperature minus lower temperature and then it is divided by the thickness okay so what we have so from the table we can have this value for k that is 0.84 [Music] joule per second per meter per degree celsius so this is the value of k the cross sectional area so 2 e 2 times 1.5 is 3 right 3 meter square so this is a and then t1 is 15 degree celsius minus 14 degree celsius that means it is one and the thickness is 3.2 millimeter so millimeter means 10 to the power minus 3 meter okay 1 millimeter means 10 to the power minus 3 meter so if you plug all these values in your calculator correctly you should get something like this seven eight eight what okay all right and this is the last one okay and this is the uh here we have to calculate the heat transfer rate by radiation okay so it says that an athlete is sitting unclothed in a locker room whose dark walls are at temperature 15 degree celsius okay so the room temperature is 15 degree celsius the person is person body temperature is 34 degree celsius emissivity is 0.70 okay and area of that person's body is 1.5 meter square okay and we have to calculate the rate of heat transfer by this athlete okay the rate of heat loss by radiation from uh i mean from this athlete so how to do that so remember that this is the equation so sigma and then t 1 to the power 4 minus t to the power 4 okay so here t1 is the body temperature and t2 is the room temperature okay so ending here so before you plug the value make sure you convert the temperature in in kelvin so 34 plus 273 and that is equal to uh that is equal to 307 kelvin t2 is 15 times plus 273 is 288 kelvin okay so that is very important many students forget to do that okay now epsilon is already given that is 0.70 and that is the stephen boltzmann's constant we can we can get the number from the textbook and that is this okay the area of the athlete's body is 1.5 meter square and then t1 is 307 to the power 4 minus okay so kelvin to the power 4 and then 288 kelvin to the power 4 okay so if you plug all these values correctly you should get you should get a value equal to 120 watt okay so the rate of heat loss from this athlete is one 120 watt okay so that's we have for the chapter okay in the next lecture i am going to start uh chapter 15 and there you will be learning about the thermodynamics laws okay so that's it for this lecture i will talk to you in the next lecture