Question

A calorimeter has mass $100$g and specific heat $0.1$ kcal/kg ${}^{o}C$. It contains $250$ gm of liquid at $30$ ${}^{o}C$ having specific heat of $0.4$ kcal/kg ${}^{o}C$. If we drop a piece of ice of mass $10$g at $0$ ${}^{o}C$. What will be the temperature of the mixture?

Answer 1

1) Net Heat Change in this Process = 0

2) ice will be converted to water, then all will come at a equilibrium. Temperature.

Units are in SI system.

Steps:

1) Given,

Mass of Caloriemeter, $m=100g=0.1kg$

Specific Heat of Caloriemeter, $S\left(C\right)=0.1units$

Mass of liquid, $m.=0.25kg$

Specific Heat of Liquid, $s\left(L\right)=0.4units$

Mass of Ice, $=0.01kg$

Latent heat of fusion of Ice, $L\left(f\right)=80units$

Specific heat of water, $S\left(w\right)=1units.$

2) Heat due to fusion $=mL$

Heat due to change in $T=ms\ast del\left(T\right)$

Let the final Equilibrium Temperature be T. Celsius.

$Q\left(net\right)=Q\left(1\right)+Q\left(2\right)+Q\left(3\right)+Q\left(4\right)0=0.1\ast 0.1(T-30)+0.25\ast 0.4(T-30)+0.01\ast 80+1\ast 0.01(T-0)$

Multiply by 100 on both sides,

$0=T-30+10(T-30)+80+T$

$12T=250$

$T=20.83°C$

Therefore, Final Temperature of mixture $=20.83°C$