Name: Carnot Cycle : Third and Fourth Stage
Uploaded: 2019-07-19
Duration: 7 min 49 s
Description: Work is done on the system during the third and fourth stages of the Carnot cycle. The video teaches us how to calculate the total work done in a Carnot cycl...

Question

A

20 g

ice cube at

- 15 ° C

is placed in a lake whose temperature is

15 ° C

. Calculate the change in the entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is

2220 J {kg}^{- 1} K^{- 1}

. (Hint: Will the ice cube affect the lake temperature?)

Accepted Answer

The ice warms to $0 ° C$ , then melts and the resulting water warms to the temperature of the lake water, which is $15 ° C .$ As the ice warms, the energy it receives as heat, when the temperature changes by $d T$ , is $d Q = m c_{1} d T$ , where $m$ is the mass of ice and $c_{1}$ is the specific heat of ice. If $T_{i} (= 258 K)$ is the initial temperature and $T_{f} (= 273 K)$ is the final temperature, then the change in its entropy is
$Δ S = \int \frac{d Q}{T} = m c_{1} \int_{T_{i}}^{T_{f}} \frac{d T}{T} = m c_{l} \ln \frac{T_{f}}{T_{i}} ∆ S = (0.020 kg) (2220 J kg K^{- 1}) \ln (\frac{273 K}{258 K}) = 2.51 J K^{- 1}$

Melting is an isothermal process. The energy leaving the ice as heat is $m L_{F}$ , where $L_{F}$ is the heat of fusion for ice. Thus,

$Δ S = \frac{Q}{T} = \frac{m L_{F}}{T} = \frac{(0.020 kg) (333 \times 10^{3} J {kg}^{- 1})}{273} = 24.40 J {kg}^{- 1}$

For the warming of the water from the melted ice, the change in entropy is
$Δ S = m c_{w} \ln \frac{T_{f}}{T_{i}}$
where $c_{w}$ is the specific heat of water $(4190 J kg K^{- 1})$ . Thus,
$Δ S = (0.020 kg) (4190 J kg K^{- 1}) \ln (\frac{288 K}{273 K}) = 4.48 J K^{- 1}$
The total change in entropy for the ice and the water,
$Δ S = 2.51 J K^{- 1} + 24.40 J K^{- 1} + 4.48 J K^{- 1} = 31.4 J K^{- 1}$

Since the temperature of the lake does not change significantly when the ice melts, the change in its entropy is $Δ S = \frac{Q}{T}$ , where $Q$ is the energy it receives as heat (the negative of the energy it supplies to the ice) and $T$ is its temperature. When the ice warms to $0 ° C$ ,
$Q = - m c_{i} (T_{f} - T_{i}) = - (0.020 kg) (2220 J kg K^{- 1}) (15 K) = - 666 J$
When the ice melts,
$Q = - m L_{F} = - (0.020 kg) (333 \times 10^{3} J {kg}^{- 1}) = - 6.66 \times 10^{3} J$
When the water from the ice warms,
$Q = - m c_{w} (T_{f} - T_{i}) = - (0.020 kg) (4190 J kg K^{- 1}) (15 K) = - 1257 J$
The total energy leaving the lake water is,
$Q = - 666 J - 6.66 \times 10^{3} J - 12.57 \times 10^{2} J = - 8.58 \times 10^{3} J$
The change in entropy is,
$Δ S = - \frac{8.58 \times 10^{3} J}{288 K} = - 29.8 J K^{- 1}$
The change in the entropy of the ice-lake system is,

$Δ S = (31.4 - 29.8) J K^{- 1} = 1.6 J K^{- 1}$

Important Questions on Entropy and the Second Law of Thermodynamics

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A 20 g ice cube at -15 °C is placed in a lake whose temperature is 15 °C. Calculate the change in the entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J kg-1 K-1. (Hint: Will the ice cube affect the lake temperature?)

Important Questions on Entropy and the Second Law of Thermodynamics

Learn and Prepare for any exam you want !