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(C) The input power $P_{	ext{input}}$ is given by $P_{	ext{input}} = V 	imes I = 100 \ V 	imes 1 \ A = 100 \ W$.Given the efficiency $\eta = 91.6\

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(C) The input power $P_{	ext{input}}$ is given by $P_{	ext{input}} = V 	imes I = 100 \ V 	imes 1 \ A = 100 \ W$.Given the efficiency $\eta = 91.6\% = 0.916$.The output power is $P_{	ext{out}} = \eta 	imes P_{	ext{input}} = 0.916 	imes 100 \ W = 91.6 \ W$.The power loss is $P_{	ext{loss}} = P_{	ext{input}} - P_{	ext{out}} = 100 \ W - 91.6 \ W = 8.4 \ W$.Since $1 \ W = 1 \ J/s$ and $1 \ cal \approx 4.2 \ J$,the power loss in $cal/s$ is $\frac{8.4 \ J/s}{4.2 \ J/cal} = 2 \ cal/s$.

$A$ motor operating on $100 \ V$ draws a current of $1 \ A$. If the efficiency of the motor is $91.6 \%$,then the loss of power in units of $cal/s$ is

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