A container of 114.0 g of water is heated using 67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from $15 ^ { \circ } \mathrm { C } \text { to } 25 ^ { \circ } \mathrm { C } \text { ? }$
The specific heat capacity of the container is negligible, and the specific heat capacity of water is $4.186 \times 10 ^ { 3 } \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C }$

Question

A container of  114.0 g of water is heated using  67.0 W of power, with perfect efficiency. How long will it take to raise the temperature of the water from $15 ^ { \circ } \mathrm { C } \text { to } 25 ^ { \circ } \mathrm { C } \text { ? }$
The specific heat capacity of the container is negligible, and the specific heat capacity of water is $4.186 \times 10 ^ { 3 } \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C }$

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 of the water in kilograms, $c$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature. Here, $m = 0.114 \, \text{kg}$, $c = 4.186 \times 10^3 \, \text{J/kg}^\circ\text{C}$, and $\Delta T = 10^\circ\text{C}$. So, $Q = 0.114 \times 4.186 \times 10^3 \times 10$. To find the time, $t$, use $Q = Pt$, where $P = 67.0 \, \text{W}$. Rearranging gives $t = Q / P$. Substituting the values gives the correct time in seconds.

A Container of 114 15∘C to 25∘C ? 15 ^ { \circ } \mathrm { C } \text { to } 25 ^ { \circ } \mathrm { C } \text { ? }15∘C to 25∘C ?

A Container of 114 $15 ^ { \circ } \mathrm { C } \text { to } 25 ^ { \circ } \mathrm { C } \text { ? }$