Question 580: Thermodynamics & KTG

Question

A water heater of power $2000 \mathrm{~W}$ is used to heat water. The specific heat capacity of water is $4200 \mathrm{~J}$ $\mathrm{kg}^{-1} \mathrm{~K}^{-1}$. The efficiency of heater is $70 \%$. Time required to heat $2 \mathrm{~kg}$ of water from $10^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ is _________ s. (Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Accepted Answer

The amount of heat energy required to raise the temperature of a substance can be calculated as:
Q = m $$ 	imes $$ c $$ 	imes $$ ÃÂT
where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ÃÂT is the change in temperature.
The time required to heat a substance can be calculated as :
t = $${Q \over P}$$
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
actual
= P
input
$$ 	imes $$ 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 kg $$ 	imes $$ 4200 J/kg/K $$ 	imes $$ (60 - 10) = 2 kg $$ 	imes $$ 4200 J/kg/K $$ 	imes $$ 50 K = 4200 $$ 	imes $$ 50 $$ 	imes $$ 2 J = 420,000 J
P
input
= 2000 W = 2000 J/s
P
actual
= 2000 $$ 	imes $$ 0.7 = 1400 J/s
t = $${Q \over {{P_{actual}}}}$$ = $${{420,000} \over {1400}}$$ J/s = 300 s
So, the time required to heat 2 kg of water from 10ÃÂ°C to 60ÃÂ°C is approximately 300 s.

Thermodynamics & KTG

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