Question

A steel ball of mass 1 kg of specific heat 0.4 kJ/kgK. is at a temperature of ${60}^{o}C$. It is dropped into 1 kg water at ${20}^{o}C$. The final steady state temperature of water is

• A. ${23.5}^{o}C$
• B. ${30}^{o}C$
• C. ${35}^{o}C$
• D. ${40}^{o}C$

Answer 1

The correct option is A ${23.5}^{o}C$

Given data:
Mass of steel ball,
$m_s=1kg$
Specific heat of steel ball,
$c_s=0.4kJ/kgK$
Initial temperature of ball,
$T_1={60}^{o}C$
Mass of water,
$m_w=1kg$
Initial temperature of water,
$T_{w1}={20}^{o}C$
Let $T_f=$ Final temperature of ball and water
Applying energy balance equation,
Heat lost by a steel ball
= Heat gained by the water
$m_s{C}_s(T_1-T_f)=m_w{C}_w(T_f-T_{w1})$
$1\times 0.4(60-T_1)=1\times 4.18(T_f-20)$
$24-0.4T_f=4.18T_f-83.6$
or $107.6=4.58T_f$
or $T_f={23.49}^{o}C\approx {23.50}^{o}C$