Question

An electric kettle or negligible heat capacity is rated at 3500 watts if 4kg of water is put into it now how long does it take the temperature of a water to rise from 30 degree celcius to 120 degree celcius . Specific heat capacity of water =4200kg-1 k-1

Answer 1

Answer on Question #72827, Physics / Molecular Physics | Thermodynamics

An electric kettle or negligible heat capacity is rated at 3500 watts if $4\mathrm{kg}$ of water is put into it now how long does it take the temperature of a water to rise from 30 degree celcius to 120 degree celcius. Specific heat capacity of water =4200kg-1 k-1

Solution

$P = \frac{Q}{t}$, where $P$ – power of heater, $Q$ – heating energy.

$Q = Cm(T_{2} - T_{1})$ – where $C$ – water specific heat capacity $(4200\mathrm{J / kg}\times \mathrm{K})$

$$t = \frac {Q}{P} = \frac {C m (T 2 - T 1)}{P};$$ $$t = \frac {4 2 0 0 \times 4 \times (1 2 0 - 3 0)}{3 5 0 0} = 4 3 2 (s)$$

Perhaps there is a mistake in a problem, because any electric kettle is not able to heat any amount of water to $120{}^{\circ}\mathrm{C}$ under pressure of 1 atm. It's impossible because water starts to evaporate. Then it turns into steam and steam leaves the kettle, otherwise the kettle will explode. Kettle cannot heat steam from $100{}^{\circ}\mathrm{C}$ to $120{}^{\circ}\mathrm{C}$. The maximum temperature is the boiling point of water: $100{}^{\circ}\mathrm{C}$.

$$t = \frac {Q}{P} = \frac {C m (T 2 - T 1)}{P};$$ $$t = \frac {4 2 0 0 \times 4 \times (1 0 0 - 3 0)}{3 5 0 0} = 3 3 6 (s)$$

Answer

432 (s) – time that kettle needs to raise the temperature of $4.0\mathrm{kg}$ of water from $30{}^{\circ}\mathrm{C}$ to $120{}^{\circ}\mathrm{C}$. (that is completely impossible).

336 (s) – time that kettle needs to raise the temperature of $4.0\mathrm{kg}$ of water from $30{}^{\circ}\mathrm{C}$ to $100{}^{\circ}\mathrm{C}$.

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