Question

A kettle contains 1.2kg of water and is supplied with energy at a rate of 2kw from the mains. Assuming the kettle is 80% effecient, how long would it take to heat the water from 20°C to the boiling point of 100°C if no energy is dissapated from the water to the environment?

Answer 1

Answer on Question#63513 - Physics - Molecular Physics

A kettle contains $1.2\mathrm{kg}$ of water and is supplied with energy at a rate of $2\mathrm{kW}$ from the mains. Assuming the kettle is $80\%$ efficient, how long would it take to heat the water from $20{}^{\circ}\mathrm{C}$ to the boiling point of $100{}^{\circ}\mathrm{C}$ if no energy is dissipated from the water to the environment?

**Solution.** The amount of heat required to heat water from temperature $20^{0}C$ to $100^{0}$ without phase change can be calculated using the formula

$$Q _ {1} = C m \left(t _ {2} - t _ {1}\right)$$

where

$C = 4184\frac{J}{kg\cdot K}$ – specific heat capacity for water

$m = 1.2\mathrm{kg}$ – mass water

$t_1 = 20^0 C$ and $t_2 = 100^0$ – the initial and final temperature, respectively.

On the other hand, the amount of heat transferred to the water, taking the efficiency of kettle can be calculated using the formula

$$Q _ {2} = \eta P t$$

where

$\eta = 0.8(80\%)$ – efficiency

$P = 2000W$ – power of kettle

$t$ – time

Because no energy is dissipated from the water to the environment get $Q_{1} = Q_{2}$.

Hence $Cm(t_{2} - t_{1}) = \eta Pt$ $t = \frac{Cm(t_{2} - t_{1})}{\eta P} = \frac{4184 \cdot 1.2(100 - 20)}{0.8 \cdot 2000} = 251\mathrm{s}$. (4 minutes 11 seconds).

**Answer.** 251s (4 minutes 11 seconds)

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