Question

In an experiment on the specific heat of a metal a 0.20 kg block of the metal at 150 ${}^\circ C$ is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 $c{{m}^{3}}$ of water at 27 ${}^\circ C$. The final temperature is 40 ${}^\circ C$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for the specific heat of the metal?

Answer 1

This is a problem of calorimetry. Metal at high temperature is dropped into a copper calorimeter containing water. The heat loss by the metal will go in increasing the temperature of the water and the calorimeter. There will be no loss of heat.

Complete step by step answer: Mass of the metal, ${{m}_{1}}$ = 0.20 kg = 200 g
The initial temperature of the metal, ${{T}_{1}}=150{}^\circ C$
The final temperature of the metal, ${{T}_{2}}=40{}^\circ C$
Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g
The volume of the water=V=150 $c{{m}^{3}}$and density of water is 1 g/$c{{m}^{3}}$.
So, using the formula $\rho =\dfrac{m}{V}$
$m=\rho V=1\times 150=150g$
We know the specific heat capacity of water is 4.186J/g/K
The rise in the temperature of the water and calorimeter system: 40-27= 13 ${}^\circ C$
Heat lost by metal= heat gained by water + heat gained by calorimeter
${{m}_{1}}c({{T}_{2}}-{{T}_{1}})=(M+m){{c}_{w}}{{T}_{w}} \\\ \implies 200\times c\times (150-40)=(150+25)\times 4.186\times 13 \\\ \implies 2000c=9523.15 \\\ \therefore c=0.43J/gK \\\$
Thus, the value of the specific heat of the metal is 0.43 J/gK. Also, If some heat is lost to the surroundings, then the value of specific heat capacity will be smaller than the actual value.

Note:
Specific heat capacity is defined as the amount of heat necessary to raise the temperature of the body by 1 ${}^\circ C$. The specific quality of a calorimeter is that it has insulating walls and its temperature also increases and thus we can find the exact amount of heat lost and gained.