Question

In an experiment on the specific heat of a metal, a $0.20\text{kg}$ block of the metal at ${150}^{\circ }C$ is dropped in a copper calorimeter (of water equivalent $0.025\text{kg}$) containing $150c{m}^3$ 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 of smaller than the actual value for specific heat of the metal.

Answer 1

Calorimeter has water equivalent of mass, $m=0.025\text{kg}=25g$

The volume of the water, $V=150c{m}^3$

Density of water is $1\frac{g}{c{m}^3}$

Step 1: Find mass of water in calorimeter.

So, using the formula $\rho =\frac{m}{v}$

$m=\rho V=1\times 150=150g$

Given, mass of the metal, $m_1=0.20\text{kg}=200g$

The initial temperature of the metal, $T_1={150}^{\circ }C$

The final temperature of the metal, $T_2={40}^{\circ }C$

Fall in temperature of the metal $T_1-T_2=(150-40{)}^{\circ }C={110}^{\circ }C$

Initial temperature of calorimeter $={27}^{\circ }C$

Final temperature of calorimeter $={40}^{\circ }C$

Step 2: Find change in temperature.

The rise in the temperature of the water and calorimeter system

$T_w=40-27={13}^{\circ }C$

Step 3: Calculate specific heat of metal.

We know that specific heat capacity of water is $C_w=4.186J/gK$

From principle of calorimetry

Heat lost by metal = heat gained by water + heat gained by calorimeter

$m_{1}c(T_2-T_1)=(M+m)C_W\Delta T$

$200\times c\times (150-40)=(150+25)\times 4.186+\times 13$

$22000c=9523.15$

$c=0.43J/gK$

The above value would be lesser than the actual value since some heat must have been lost to the surroundings as well which we haven't accounted for.