Question

670 meter Cubic Cupper ball at 104°C is thrown into a lake which is at 19°C determine the amount of heat loss by the ball after it reaches a thermal equilibrium with the lake ,base your calculations on the room temperature

The answer should be -195.6 kj

Please explain each stepin words in details

Thank you

Answer 1

Solution .

given informations

volume of copper ball V=670 cm^3 (not be m^3 as given otherwise it would be too huge)

=670*10^-6 m^3

temprature Tc=104 C

lake temprature Tl= 19 C

properties of copper:

density =8960 kg/m^3

specific heat c=.385 kJ/kg-k

thus mass of copper ball mc=*V

=8960*670*10^-6

=6.0032 kg

Lake is infinite body its temprature remains same for a small amount of heat abosorption thus thermal equillibrium temprature would be 19 C

According to energy conservation, heat loss=heat gain

heat loss by copper ball=mc*c*(Ti-Tf)

=6.0032*.385*(104-19)

=196.41545 kJ

As i considered heat loss that why i didnot take negative sign in account otherwise it would be -196.41545 kJ

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