A metal piece of thermal capacity 40 JK -1 , absorbs 800 J of heat. Calculate the rise in the temperature of this metal piece.

Given,

• Thermal capacity of the metal piece = 40 JK^-1
• Heat absorbed by the substance = 800 J

Let, rise in temperature be T.

As, thermal capacity of a substance is given by,

$$\begin{aligned}\text {Thermal capacity} = \frac{\text {Heat absorbed}}{\text {Temperature rise}} \\[1em] \Rightarrow \text {Temperature rise} = \frac{\text {Heat absorbed}}{\text {Thermal capacity}} \\[1em] \Rightarrow \text T = \frac{800}{40} \\[1em] \Rightarrow \text T = 20\ \text K\end{aligned}$$

Since a rise of 1 K = 1°C, the rise in temperature is also 20°C.

Hence, the rise in the temperature of the metal piece is 20°C.