Question

A 33.0 g iron rod, initially at 22.7 ∘c, is submerged into an unknown mass of water at 63.3 ∘c, in an insulated container. the final temperature of the mixture upon reaching thermal equilibrium is 58.5 ∘c. part a what is the mass of the water? express your answer to two significant figures and include the appropriate units.

Answer 1

When the iron and the water reach thermal equilibrium, they have same temperature, $T=58.5^{\circ}C$.
We can consider this as an isolated system, so the heat released by the water is equal to the heat absorbed by the iron.

The hear released by the water is:
$Q_w =m_w C_{sw} \Delta T_w$
where $m_w$ is the water mass, $C_{sw}=4.186 J/(g^{\circ}C)$ is the specific heat of the water, and $\Delta T_w = 63.6^{\circ}-58.5^{\circ}=5.1^{\circ}C$ is the variation of temperature of the water.

Similarly, the heat absorbed by the iron is:
$Q_i = m_i C_{si} \Delta T_i$
where $m_i = 33.0 g$ is the iron mass, $C_{si}=0.444 J/(g^{\circ}C)$ is the iron specific heat, and $\Delta T_i = 58.5^{\circ}-22.7^{\circ}=35.8^{\circ}C$ is the variation of temperature of the iron.

Writing $Q_w=Q_i$ and replacing the numbers, we can solve to find mw, the mass of the water:
$m_w= \frac{m_i C_{si} \Delta T_i}{C_{sw} \Delta T_w} =24.6 g$