Worked examples · Sheet 2

Heat balances with changes of state

Theory used: Latent heat Heat capacity and specific heat Heat balances with changes of state Equilibrium temperature

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the heat needed to turn $0.200 kg$ of ice at $- 30.0^{\circ} C$ into water at $40.0^{\circ} C$ , and draw the temperature as a function of the heat absorbed. Specific heats: $c_{g} = 2090 J/ (kg \cdot K)$ for ice, $c_{a} = 4186 J/ (kg \cdot K)$ for water; latent heat of fusion $λ_{f} = 3.33 \cdot 1 0^{5} J/kg$ .

Q = Q_{1} m c_{g} (0 - T_{i}) + Q_{2} m λ_{f} + Q_{3} m c_{a} (T_{f} - 0)

Q_{1} = 0.200 \cdot 2090 \cdot 30.0 J = 1.25 \cdot 1 0^{4} J

Q_{2} = 0.200 \cdot 3.33 \cdot 1 0^{5} J = 6.66 \cdot 1 0^{4} J

Q_{3} = 0.200 \cdot 4186 \cdot 40.0 J = 3.35 \cdot 1 0^{4} J

Q = (1.25 + 6.66 + 3.35) \cdot 1 0^{4} J \approx 1.13 \cdot 1 0^{5} J

Temperature of the sample as a function of the heat absorbed, with

Q

1 0^{4} J

; during melting the temperature stays at

0^{\circ} C

The process is divided into three stages, because the law $Q = m c Δ T$ holds only as long as the substance does not change state. In the first stage the ice warms up to $0^{\circ} C$ with the specific heat of ice; in the second it melts at constant temperature and absorbs the latent heat $m λ_{f}$ ; in the third the liquid water warms up to $40^{\circ} C$ with the specific heat of water. The total heat is the sum of the three terms, all positive because the sample absorbs heat in every stage.

The graph shows that melting requires more heat than the two heating stages put together: on the horizontal segment the heat absorbed does not change the temperature. The slopes of the inclined segments are $\frac{1}{m c}$ , therefore the segment of the ice is about twice as steep as that of the water, since $c_{g}$ is about half of $c_{a}$ .

Exercise 2

$0.0500 kg$ of ice at $0^{\circ} C$ is placed in $0.400 kg$ of water at $30.0^{\circ} C$ . The container has negligible heat capacity and is insulated from the outside. Determine the final state and the equilibrium temperature ( $c_{a} = 4186 J/ (kg \cdot K)$ , $λ_{f} = 3.33 \cdot 1 0^{5} J/kg$ ).

Q_{avail} = m_{a} c_{a} (T_{a} - 0) = 0.400 \cdot 4186 \cdot 30.0 J = 5.02 \cdot 1 0^{4} J

Q_{melt} = m_{g} λ_{f} = 0.0500 \cdot 3.33 \cdot 1 0^{5} J = 1.67 \cdot 1 0^{4} J

Q_{avail} > Q_{melt} ⟹ all the ice melts

m_{g} λ_{f} + m_{g} c_{a} (T_{e} - 0) + m_{a} c_{a} (T_{e} - T_{a}) = 0

T_{e} = \frac{m _{a} c _{a} T _{a} - m _{g} λ _{f}}{( m _{a} + m _{g} ) c _{a}}

T_{e} = \frac{50 232 - 16 650}{0.450 \cdot 4186}^{\circ} C = \frac{33 582}{1883.7}^{\circ} C \approx 17.8^{\circ} C

Before writing the balance, we establish whether all the ice melts. The heat that the water would release by cooling down to $0^{\circ} C$ is greater than the heat needed to melt the ice, therefore the ice melts completely and the final temperature is above $0^{\circ} C$ .

Three terms appear in the balance: the melting of the ice, the warming from $0^{\circ} C$ to $T_{e}$ of the water that comes from the ice, with the specific heat of water, and the cooling of the initial water. Their sum is zero because the system is isolated. The equation is linear in $T_{e}$ and gives $T_{e} \approx 17.8^{\circ} C$ . As a check, without melting the equilibrium temperature would be the weighted average $\frac{0.400 \cdot 30.0}{0.450}^{\circ} C \approx 26.7^{\circ} C$ ; the value found is lower because part of the heat of the water was used to melt the ice.

Exercise 3

$0.300 kg$ of ice at $- 10.0^{\circ} C$ is placed in $0.500 kg$ of water at $20.0^{\circ} C$ . The container has negligible heat capacity and is insulated from the outside. Determine the final temperature and the mass of ice that melts ( $c_{a} = 4186 J/ (kg \cdot K)$ , $c_{g} = 2090 J/ (kg \cdot K)$ , $λ_{f} = 3.33 \cdot 1 0^{5} J/kg$ ).

Q_{avail} = m_{a} c_{a} (T_{a} - 0) = 0.500 \cdot 4186 \cdot 20.0 J = 4.19 \cdot 1 0^{4} J

Q_{warm} = m_{g} c_{g} (0 - T_{g}) = 0.300 \cdot 2090 \cdot 10.0 J = 6.27 \cdot 1 0^{3} J

Q_{melt} = Q_{warm} + m_{g} λ_{f} = (6270 + 99900) J = 1.06 \cdot 1 0^{5} J

Q_{warm} < Q_{avail} < Q_{melt} ⟹ T_{e} = 0^{\circ} C

m_{melted} = \frac{Q _{avail} - Q _{warm}}{λ _{f}} = \frac{41 860 - 6270}{3.33 \cdot 1 0 ^{5}} kg \approx 0.107 kg

The heats are compared taking $0^{\circ} C$ as the reference. The heat that the water can release by cooling down to $0^{\circ} C$ is enough to warm the ice up to $0^{\circ} C$ , but it is less than the heat that would be needed to melt all of it. Therefore the ice melts only partially and the final state is a mixture of water and ice, which at atmospheric pressure can be in equilibrium only at $T_{e} = 0^{\circ} C$ .

With $T_{e} = 0^{\circ} C$ the water releases exactly $Q_{avail}$ . Part of this heat brings the ice from $- 10^{\circ} C$ to $0^{\circ} C$ , and the rest melts the mass $m_{melted}$ , which is obtained by dividing by the latent heat. At the end $0.193 kg$ of ice and $0.607 kg$ of water remain, both at $0^{\circ} C$ . Writing the balance immediately with $T_{e}$ unknown and complete melting would give a negative temperature, incompatible with the hypothesis that all the ice has melted.

Exercise 4

$0.0200 kg$ of water vapor at $100^{\circ} C$ is introduced into a container holding $0.500 kg$ of water at $15.0^{\circ} C$ . The container has negligible heat capacity and is insulated from the outside. Compute the equilibrium temperature ( $c_{a} = 4186 J/ (kg \cdot K)$ , $λ_{v} = 2.26 \cdot 1 0^{6} J/kg$ ).

Q_{cond} = m_{v} λ_{v} = 0.0200 \cdot 2.26 \cdot 1 0^{6} J = 4.52 \cdot 1 0^{4} J

m_{a} c_{a} (100 - T_{a}) = 0.500 \cdot 4186 \cdot 85.0 J = 1.78 \cdot 1 0^{5} J > Q_{cond}

- m_{v} λ_{v} + m_{v} c_{a} (T_{e} - 100) + m_{a} c_{a} (T_{e} - T_{a}) = 0

T_{e} = \frac{m _{v} λ _{v} + m _{v} c _{a} \cdot 100 + m _{a} c _{a} T _{a}}{( m _{v} + m _{a} ) c _{a}}

T_{e} = \frac{45 200 + 8372 + 31 395}{0.520 \cdot 4186}^{\circ} C = \frac{84 967}{2176.7}^{\circ} C \approx 39.0^{\circ} C

The vapor condenses and releases the latent heat, then the condensed water cools from $100^{\circ} C$ to $T_{e}$ ; the initial water absorbs all this heat. The first check verifies that the condensation is complete: even by bringing the water up to $100^{\circ} C$ one could absorb much more heat than the heat of condensation, therefore all the vapor condenses and $T_{e}$ is below $100^{\circ} C$ .

In the balance the condensation appears with a minus sign, because it is a heat released. One obtains $T_{e} \approx 39.0^{\circ} C$ . Just $20 g$ of steam warm half a kilogram of water by $24 K$ ; liquid water at $100^{\circ} C$ of the same mass would bring it only to about $18^{\circ} C$ , because the latent heat of vaporization is about five times the heat needed to warm water from $0^{\circ} C$ to $100^{\circ} C$ . For this reason a steam burn is more serious than a burn from boiling water.