Specific Heat Capacity Problem Solving Example

Problem Statement

• Calculate the energy required to heat 20 grams of water.
• Initial temperature: 283°C
• Final temperature: 303°C
• Specific heat capacity of water (C): 4.18 J/g°C

Formula

• Q = mcΔT
  • Q: Energy absorbed (in Joules)
  • m: Mass of the substance (in grams)
  • c: Specific heat capacity (J/g°C)
  • ΔT: Change in temperature (°C)

Steps to Solve

1. Identify What Needs to be Solved

   • Solve for Q (Energy absorbed).

2. Gather Information

   • Mass (m) = 20 grams.
   • Specific heat capacity (c) = 4.18 J/g°C.
   • Initial Temperature (T_initial) = 283°C.
   • Final Temperature (T_final) = 303°C.

3. Calculate Temperature Change (ΔT)

   • ΔT = T_final - T_initial
   • ΔT = 303°C - 283°C = 20°C

4. Compute Energy (Q)

   • Q = m × c × ΔT
   • Q = 20 g × 4.18 J/g°C × 20°C
   • Q = 1672 Joules

5. Significant Figures

   • Round the answer to three significant figures: 1670 Joules

Conclusion

• The energy needed to increase the temperature of 20 grams of water from 283°C to 303°C is 1670 Joules when rounded to three significant figures.