A container of 114.0 g of water is heated using $67.0 \mathrm {~W}$ of power, with perfect efficiency. How long will it take to raise the temperature of the water from $15 ^ { \circ } \mathrm { C }$ to $25 ^ { \circ } \mathrm { C } \text { ? }$ The specific heat capacity of the container is negligible, and the specific heat capacity of water is 4.186 × 10³ J/kg ∙ C.

Question

Quizplus · Accepted Answer

To find the time, use the formula $ Q = mc\Delta T $, where $ m = 0.114 \, \text{kg} $, $ c = 4186 \, \text{J/kg} \cdot \text{C} $, and $ \Delta T = 10 \, \text{C} $. Calculate $ Q = 0.114 \times 4186 \times 10 = 4772.04 \, \text{J} $. Power $ P = 67 \, \text{W} $ means $ 67 \, \text{J/s} $. Time $ t = \frac{Q}{P} = \frac{4772.04}{67} \approx 71 \, \text{s} $.

A Container of 114 67.0 W67.0 \mathrm {~W}67.0 W Of Power, with Perfect Efficiency

A Container of 114 $67.0 \mathrm {~W}$ Of Power, with Perfect Efficiency