A small electric immersion heater is used to heat $\mathbf{\text{100g}}$of water for a cup of instant coffee. The heater is labeled “$\mathbf{200}\mathbf{}\mathbf{watts}$” (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from$\mathbf{23.0}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{to}\mathbf{}\mathbf{100}\mathbf{°}\mathbf{C}$, ignoring any heat losses.

1. The mass of the water is$\mathrm{m}=100\mathrm{g}\text{or}100\times {10}^{-3}\mathrm{kg}$
2. The electric power of the heater is .$\mathrm{P}=200\mathrm{W}$.
3. The initial temperature of the water is${\mathrm{T}}_{\mathrm{i}}=23°\mathrm{C}\text{or}296\mathrm{K}$
4. The final temperature of the water is${\mathrm{T}}_{\mathrm{f}}=100°\mathrm{C}\text{or}373\mathrm{K}$
5. The specific heat of water, $\mathrm{c}=4190\mathrm{J}/\mathrm{kg}.\mathrm{K}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of the specific heat of the water. We can use the definition of power to find the time required for the water to boil.

Formulae:

The heat energy required by a body, $\mathrm{Q}=\mathrm{mc\Delta T}$ …(i)

Where, $\mathrm{m}$= mass

$\mathrm{c}$= specific heat capacity

$∆T$= change in temperature

$Q$= required heat energy

The heat power generated by a body, $P=\frac{Q}{t}$ …(ii)

Using equation (i) in equation (ii) and substituting the given values, we get the time required to bring the water temperature from$23°\mathrm{C}$ to$100°\mathrm{C}$is given as:

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{cm}({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}})}{\mathrm{P}}\\ =\frac{4190\mathrm{J}/\mathrm{kg}.\mathrm{K}\times 100\times {10}^{-3}\mathrm{kg}\times (373\mathrm{K}-296\mathrm{K})}{200\mathrm{W}}\\ =160\mathrm{s}\end{array}$

Hence, the required time is$160\mathrm{s}$