Thermodynamics Part 4

The amount of heat energy required to raise the temperature of a substance can be calculated as:$Q = m \times c \times \Delta T$where $Q$ is the heat energy required, $m$ is the mass of the substance, $c$ is its specific heat capacity, and $\Delta T$ is the change in temperature.The time required to heat a substance can be calculated as :$t = \frac{Q}{P}$where $t$ is the time required, and $P$ is the power of the heating device.The actual power output of the heating device can be calculated as:$P_{\text{actual}} = P_{\text{input}} \times \text{efficiency}$where $P_{-}$input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.Substituting the given values:$Q = 2\,\text{kg} \times \frac{4200\,\text{J}}{\text{kg}\,\text{K}} \times (60-10) = 2\,\text{kg} \times \frac{4200\,\text{J}}{\text{kg}\,\text{K}} \times 50\,\text{K} = 4200 \times 50 \times 2\,\text{J} = 420,\!000\,\text{J}$ $P_{\text{input}} = 2000\,\text{W} = \frac{2000\,\text{J}}{\text{s}}$$P_{\text{actual}} = 2000 \times 0.7 = \frac{1400\,\text{J}}{\text{s}}$$t = \frac{Q}{P_{\text{actual}}} = \frac{420,000}{1400}\,\frac{\text{J}}{\text{s}} = 300\,\text{s}$So, the time required to heat $2\,\text{kg}$ of water from $10^{\circ}\,\text{C}$ to $60^{\circ}\,\text{C}$ is approximately $300\,\text{s}$.