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A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase when heat is transferred to the oil at a rate of 1 W and 1.5 W of power is applied to the stirring device.

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To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.

Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,


\dot{U} = \dot{Q}-\dot{W}

Substitute 1W for
\dot{Q} and 1.5 W for
\dot{W}


\dot{U} = 1-1(1.5)


\dot{U} = 2.5W

Now calculcate the rate of specific internal energy increase,


\dot{u} = \frac{\dot{U}}{m}


\dot{u} = (2.5)/(1.5)


\do{u} = 1.6667W/kg

The rate of specific internal energy increase is 1.6667W/kg

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