Question

A 215 g lead ball at a temperature of 83.1 ∘C is placed in a light calorimeter containing 158 g of water at 21.3 ∘C. Find the equilibrium temperature of the system.

Answer 1

Solution:

Given;

mass of lead $m_{m}=215g$

mass of water $m_{w}=158g$

Temperature of lead

Tm = 83.1°C

Temperature of water $T_{w}=21.3^0C$

Specific heat capacity of lead $c_{m}=0.0305cal/g^0C$

Specific heat capacity of water $c_{w}=1cal/g^0C$

Let the final temperature of the system be $T_{f}$ .

By the principle of calorimetery , we have;

  

  $215\times0.0305\times(83.1-T_{f}) =158\times1\times(T_{f}-21.3)$

  $6.5575\times(83.1-T_{f}) =158\times(T_{f}-21.3)$

  

$T_{f} =\frac{3910.32825}{164.5575}=23.76^0C$

The equilibrium temperature of the system is $23.76^0C$ .