Question

A 0.20 kg block of an unknown substance at 75°C is placed in 0.5kg of water with temperature 20°C . The final temperature of the system is 20.68°C. What is the specific heat of the substance?
mA = 0.20 kg C A = ?, T A = 75°C, m B = 0.50 kg, C B = 4180 J/(kg·k), T B = 21°C Tf=20.68

Answer 1

Given:-

mass of block, $m_{b}$=0.2 kg

Initial temperature of block, $T_{bi}=75^{0}C$

mass of water, $m_{w}=0.5kg$

Initial temperature of water, $T_{wi}=20^{0}C$

Final temperature of the system, $T_{f}=20.68^{0}C$

specific heat of water, $C_{w}=4.18kJ/kg-K$

To Find:-

specific heat of the block

Explanation:-

Let specific heat of the block be $C_{b}$

From Energy conservation, heat gained by water=heat lost by block

$\Rightarrow m_{b}\times C_{b}\times(T_{bi}-T_{f})=m_{w}\times C_{w}\times (T-T_{wi})\\\\\Rightarrow 0.2\times C_{b}\times(75-20.68)=0.5\times4.18\times(20.68-20)\\\\\Rightarrow C_{b}=0.1308kJ/kg-K$