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

6.0 g

ice cube at

- 10 ° C

is put into a thermos flask containing

100 {cm}^{3}

of water at

20 ° C

. By how much has the entropy of the cube-water system changed when equilibrium is reached? The specific heat of ice is

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

.

Accepted Answer

In coming to equilibrium, the heat lost by $100 {cm}^{3}$ of liquid water (of mass $m_{w} =$ $100 g$ and specific heat capacity $c_{w} = 4190 J kg K^{- 1}$ ) is absorbed by the ice (of mass $m_{i}$ ), which melts and reaches $T_{f} > 0 ° C$ . We begin by finding the equilibrium temperature,
$\sum Q = 0$

$Q_{warm water cools} + Q_{ice warms to 0 °} + Q_{ice melts} + Q_{melted ice warms} = 0$

$c_{w} m_{w} (T_{f} - 20 °) + c_{i} m_{i} (0 ° - (- 10 °)) + L_{F} m_{i} + c_{w} m_{i} (T_{f} - 0 °) = 0$
which yields, after using $20 ° C$ and the values cited in the problem, $T_{f} = 12.24 ° C$ , which is equivalent to $T_{f} = 285.39 K$ .

In an isochoric process, change in entropy is given by

$∆ Q = \int \frac{d Q}{T} = \int \frac{d U}{T} = m c \int \frac{d T}{T}$

"Entropy change of two blocks coming to equilibrium" shows that,
$Δ S_{temp change} = m c \ln (\frac{T_{2}}{T_{1}})$
For processes where $Δ T = T_{2} - T_{1}$ ,

During melting,

$Δ S_{melt} = \frac{L_{F} m}{T_{0}}$ for the phase change experienced by ice (with $T_{0} = 273.15 K$ )

The total entropy change is (with $T$ in $kelvin$ )
$Δ S_{system} = m_{w} c_{w} \ln (\frac{285.39}{293.15}) + m_{i} c_{i} \ln (\frac{273.15}{263.15}) + m_{i} c_{w} \ln (\frac{285.39}{273.15}) + \frac{L_{F} m_{i}}{273.15}$

$= (- 11.24 + 0.66 + 1.47 + 9.75) J K^{- 1} = 0.64 J K^{- 1}$

A $6.0 g$ ice cube at $- 10 ° C$ is put into a thermos flask containing $100 {cm}^{3}$ of water at $20 ° C$ . By how much has the entropy of the cube-water system changed when equilibrium is reached? The specific heat of ice is $2220 J {kg}^{- 1} K^{- 1}$ .

Important Questions on Entropy and the Second Law of Thermodynamics

Learn and Prepare for any exam you want !

A 6.0 g ice cube at -10 °C is put into a thermos flask containing 100 cm3 of water at 20 °C. By how much has the entropy of the cube-water system changed when equilibrium is reached? The specific heat of ice is 2220 J kg-1 K-1.

Important Questions on Entropy and the Second Law of Thermodynamics

Learn and Prepare for any exam you want !

A $6.0 g$ ice cube at $- 10 ° C$ is put into a thermos flask containing $100 {cm}^{3}$ of water at $20 ° C$ . By how much has the entropy of the cube-water system changed when equilibrium is reached? The specific heat of ice is $2220 J {kg}^{- 1} K^{- 1}$ .