Question

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.

Answer 1

Heat transferred through the battery is given by

$Heat = Power \times time$

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

so here we will have

$P = V \times i$

here we will have

we know that

$V = 12 volts$

$i = 5 A$

now we will have

$P = 12\times 5 = 60 watt$

now this system is used for t = 5 min

so total energy transferred is given by

$E = P\times t$

$E = 60 \times (5\times 60)$

$E = 18000 J$

now we have

$E = 18 kJ$

so the energy would be 18 kJ