Respuesta :
Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation
[tex]\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}[/tex]
[tex]Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}[/tex]
QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
The amount of energy that must be supplied to the device is calculated using the opposite principle of a heat engine. The value obtained is 2,359,104.39 J.
Refrigerator and Carnot Engine
Given that cooler temperature is,
[tex]T_1=0^\circ \,C = 273\,K[/tex]
The hotter temperature is;
[tex]T_2=22.5^\circ \,C = 295.5\,K[/tex]
The heat used to melt a certain mass of water can be given as;
[tex]Q_1 = m\times L_f[/tex]
Substituting the known values, we get;
[tex]Q_1 = 85.7\,kg \times 3.34\times 10^5\,J/kg = 28623800\,J[/tex]
Now, we use the Carnot engine equation, we get;
[tex]\frac{Q_1}{Q_2} =\frac{T_1}{T_2}[/tex]
Therefore,
[tex]Q _2 = \frac{Q_1 T_2}{T_1}=\frac{28623800\,J \times295.5\,K }{273\,K} =30982904.39\,J[/tex]
The work on the refrigerator is given by;
[tex]W= Q_2-Q_1= 30982904.39\,J-28623800\,J=2359104.39\,J[/tex]
Hence the energy that must be applied is [tex]E=2359104.39\,J[/tex]
Learn more about Carnot engine here;
https://brainly.com/question/14983940
