A 10 g piece of iron $\left(C=0.45 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\right)$ at $100^{\cir

Question

A 10 g piece of iron $\left(C=0.45 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\right)$ at $100^{\circ} \mathrm{C}$ is dropped into 25 g of water $\left(C=4.2 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\right)$ at $27^{\circ} \mathrm{C}$. Find temperature of the iron and water system at thermal equilibrium.

Professor PC Thomas Classes · Answer

(a) Heat lost by iron + Heat gained by water $=0$
 $ \begin{gathered} 10 	imes 0.45(T-373)+25 	imes 4.2 	imes(T-300)=0 \ T=303 \mathrm{~K} 	ext { or } 30^{\circ} \mathrm{C} \end{gathered} $