Question

A motor supplies $0.4$ hp to stir 5 kg of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water $6{ }^{\circ} \mathrm{C}$ ? The heat required to heat the water is $$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6{ }^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$ This is actually supplied by friction work, so Friction work done $=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}$ and this equals the work done by the motor. But Work done by motor in time $t=($ Power $)(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)$. Equating this to our previous value for the work done yields $$ t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min} $$

          A motor supplies $0.4$ hp to stir 5 kg of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water $6{ }^{\circ} \mathrm{C}$ ?
The heat required to heat the water is
$$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6{ }^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$
This is actually supplied by friction work, so
Friction work done $=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}$
and this equals the work done by the motor. But Work done by motor in time $t=($ Power $)(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)$.
Equating this to our previous value for the work done yields
$$
t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min}
$$

Added by Lisa J.

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