Question

A $10g$ peice of iron $(C=0.45J/g^{o}C)$ at ${100}^{o}C$ is dropped into $25g$ of water $(C=4.2J/g^{o}C)$ at ${27}^{o}C.$ Find temperature of the iron and water system at thermal equilibrium :

• A. ${30}^{o}C$
• B. ${33}^{o}C$
• C. ${40}^{o}C$
• D. None of these

Answer 1

The correct option is A ${30}^{o}C$

Given mass of iron$=10g$; Specific heat(C)$=0.45J/g^{\circ }C$; Temperature$={100}^{\circ }C$ or $373K$

Given mass of water $=25g$; Specific heat(C)$=4.2J/g^{\circ }C$; Temperature$={27}^{\circ }C$ or $300K$

At thermal equilibrium;

Heat lost by iron = Heat gained by water (since heat will flow from iron to water)

$mC\Delta T_{iron}=mC\Delta T_{water}$

$\Rightarrow 10\times 0.45\left(T_{EQ}-100\right)=25\times 4.2\left(T_{EQ}-27\right)$

$\Rightarrow T_{EQ}={30}^{\circ }C$