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= (m_i C_(si) \Delta T_i)/(C_(sw) \Delta T_w) =24.6 g`