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 of the container is negligible,and the specific heat of water is 4.186 × 10³ J/kg ∙ C.

Question

Quizplus · Accepted Answer

To calculate the time required to heat the water, use the formula $Q = mc\Delta T$, where $Q$ is the heat energy in joules, $m$ is the mass in kilograms, $c$ is the specific heat capacity (4.186 J/g°C for water), and $\Delta T$ is the change in temperature. Here, $m = 114.0 \, \text{g} = 0.114 \, \text{kg}$, $c = 4.186 \, \text{J/g°C}$, and $\Delta T = 25°C - 15°C = 10°C$. Thus, $Q = 0.114 \, \text{kg} \times 4.186 \, \text{J/g°C} \times 10°C = 477.084 \, \text{J}$. Since power ($P$) is the rate of energy transfer, and $P = 67.0 \, \text{W} = 67.0 \, \text{J/s}$, the time ($t$) can be found by $t = \frac{Q}{P} = \frac{477.084 \, \text{J}}{67.0 \, \text{J/s}} \approx 7.1 \, \text{s}$.

A Container of 114 67.0 W67.0 \mathrm {~W}67.0 W Of Power,with Perfect Efficiency 15∘C15 ^ { \circ } \mathrm { C }15∘C

A Container of 114 $67.0 \mathrm {~W}$ Of Power,with Perfect Efficiency $15 ^ { \circ } \mathrm { C }$