Question

A 10 g piece of iron $(C=0.45J/g^{0}C)$ at ${100}^{0}C$ is dropped in to $25g$ of water $(C=4.2J/g^{0}C)$ at ${27}^{0}C$ . Find temperature of the iron and water system at thermal equilibrium.

• A. ${30}^{0}C$
• B. ${27}^{0}C$
• C. ${40}^{0}C$
• D. None of these

Answer 1

The correct option is D None of these

Let T be temperature at equilibrium,

According to the principle of calorimetry,

Heat gained by water = Heat lost by iron

$\Rightarrow mc\Delta T=mc\Delta T$

$\Rightarrow 25\times 4.2\times (T-27)=10\times 0.45\times (T-100)$

$\Rightarrow 105T-2835=4.5T-450$

$\Rightarrow 100.5T=2385$

$\Rightarrow T=23.7°C$

Hence, the correct option is D.