Question

Some small aluminum riverts of total mass 170g and at 100°C are emptied into a hole in a large block of ice at 0°C.
a) What will be the final temperature of the riverts?
b) How much ice will melt?

Answer 1

Some small aluminum rivets of total mass 170g and at $100{}^{\circ}\mathrm{C}$ are emptied into a hole in a large block of ice at $0{}^{\circ}\mathrm{C}$.

a) What will be the final temperature of the rivets?

b) How much ice will melt?

Please explain your answer

Specific heat of aluminum: $c_{a} = 930J / kg * K$, and heat of fusion of ice is: $\lambda_{i} = 335 * 10^{3}J / kg$

As we have a large block of ice, the final temperature of the rivet is equal to the ice temperature $= 0{}^{\circ}\mathrm{C}$.

The rivet will give to ice energy:

$$Q = m _ {a} c _ {a} \left(T _ {2} - T _ {1}\right)$$

This heat will melt ice:

$$Q = m _ {i} \lambda_ {i} \rightarrow m _ {i} = \frac {Q}{\lambda_ {i}}$$ $$m _ {i} = \frac {m _ {a} c _ {a} (T _ {2} - T _ {1})}{\lambda_ {i}} = \frac {0 . 1 7 k g * 9 3 0 J / k g \cdot K * 1 0 0 K}{3 3 5 * 1 0 ^ {3} J / k g} = 0. 0 4 7 k g$$

Answer: $T_{2} = 0{}^{\circ}\mathrm{C}, m_{i} = 0.047\mathrm{kg}$