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 $25 ^ { \circ } \mathrm { C } ?$ The specific heat capacity of the container is negligible, and the specific heat capacity of water is 4.186 × 10³ J/kg ? C. A)71 s B)4.1 s C)17 s D)320,000 s

Question

Quizplus · Accepted Answer

Explanation:The heat required to raise the temperature of the water is:$$Q = mc_p \Delta T = (114.0 \mathrm {~g})(4.186 	imes 10^3 \mathrm {~J/kg \cdot C})(25 ^ { \circ } \mathrm { C } - 15 ^ { \circ } \mathrm { C }) = 4.79 	imes 10^4 \mathrm {~J}$$The time required to heat the water is:$$t = \frac{Q}{P} = \frac{4.79 	imes 10^4 \mathrm {~J}}{67.0 \mathrm {~W}} = 71.5 \mathrm {~s}$$Therefore, the correct answer is A) 71 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