Question

A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is added to the water and ice in the container. what is the final temperature of the system? (specific heat of ice = 2,000 j/kg°c , specific heat of water = 4,186 j/kg°c, heat of fusion of water = 333.7 kj/kg, specific heat of glass = 837.2 j/km°c, specific heat of lead = 127.7 j/km°c)

Answer 1

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

• $E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
• $m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$. Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

• $Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
• $Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
• $Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$.

$E(\text{absorbed} ) = E(\text{released})$

• $E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
• $E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$:

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$. However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.