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As shown in fig. p3.64, a closed, rigid tank fitted with a fine-wire electric resistor is filled with refrigerant 22, initially at 2108c, a quality of 80%, and a volume of 0.01 m3. a 12-volt battery provides a 5-amp current to the resistor for 5 minutes. if the final temperature of the refrigerant is 408c, determine the heat transfer, in kj, from the refrigerant.

Respuesta :

Heat transferred through the battery is given by

[tex]Heat = Power \times time[/tex]

now we know that power due to battery is given by product of voltage and current

so here we will have

[tex]P = V \times i[/tex]

here we will have

we know that

[tex]V = 12 volts[/tex]

[tex] i = 5 A[/tex]

now we will have

[tex]P = 12\times 5 = 60 watt[/tex]

now this system is used for t = 5 min

so total energy transferred is given by

[tex]E = P\times t[/tex]

[tex]E = 60 \times (5\times 60)[/tex]

[tex]E = 18000 J[/tex]

now we have

[tex]E = 18 kJ[/tex]

so the energy would be 18 kJ

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