Question

A metal piece of thermal capacity 40 JK\(^{-1}\), absorbs 800 J of heat. Calculate the rise in the temperature of this metal piece.

Hint:

Distinguish between specific heat capacity (per unit mass, \(c\)) and thermal capacity (for the whole object, \(C'\)). The formulas are \(Q=mc\Delta T\) and \(Q=C'\Delta T\). Here, the problem gives thermal capacity directly, making the calculation simpler.

Answer 1

Step 1: Identify the given quantities:
- Heat absorbed (\(Q\)) = 800 J
- Thermal capacity (\(C'\)) = 40 J K\(^{-1}\)
Step 2: Recall the relevant formula:
The relationship between heat absorbed (\(Q\)), thermal capacity (\(C'\)), and the change in temperature (\(\Delta T\)) is given by:
\[ Q = C' \times \Delta T \] Step 3: Rearrange the formula to solve for the rise in temperature (\(\Delta T\)):
\[ \Delta T = \frac{Q}{C'} \] Step 4: Substitute the given values and calculate:
\[ \Delta T = \frac{800 \text{ J}}{40 \text{ J K}^{-1}} \] \[ \Delta T = 20 \text{ K} \] Since a change of 1 K is equal to a change of 1\(^{\circ}\)C, the rise in temperature can also be expressed as 20\(^{\circ}\)C.