If 69.5 kJ of heat is applied to a 1012-g block of metal, the temperature of the metal increases by 11.4°C. What is the specific heat capacity of the metal in J/g°C?

Question

Wyatt Atkins · Accepted Answer

#=6.024J//g.^@C# #W=mcDeltaT# #therefore c= W/(mDeltaT)# #=(69500J)/((1012g)(11.4^@C)# #=6.024J//g.^@C#. (Note that the SI unit of specific heat capacity is actually #J//kg.K#).

Avery Adams · Answer

undefined The specific heat capacity of the metal is 0.646 J/g°C.

HIX Tutor · Answer

undefined Let’s find the specific heat capacity step by step.

1. **Heat Energy (Q)**: You have 69.5 kJ. How can we change this to joules?  
   - 1 kJ = 1000 J, so:
   - $ 69.5 \text{ kJ} = 69.5 \times 1000 \text{ J} = 69,500 \text{ J} $

2. **Mass (m)**: The mass of the metal is 1012 g.

3. **Temperature Change (ΔT)**: The temperature increase is 11.4°C.

4. **Formula**: The formula for specific heat capacity (c) is:
   $$
   c = \frac{Q}{m \times ΔT}
   $$

5. **Plug in the numbers**: Now substitute into the formula:
   $$
   c = \frac{69,500 \text{ J}}{1012 \text{ g} \times 11.4°C}
   $$

6. **Calculate**: First, multiply the mass by the temperature change:
   $$
   1012 \times 11.4 = 11,548.8
   $$

7. **Final Calculation**:
   $$
   c = \frac{69,500 \text{ J}}{11,548.8} \approx 6.02 \text{ J/g°C}
   $$

So, what did we find for the specific heat capacity of the metal?