JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 19)
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).
Explanation
The amount of heat energy required to raise the temperature of a substance can be calculated as:
Q = m $$ \times $$ c $$ \times $$ ΔT
where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT 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:
Pactual = Pinput $$ \times $$ 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 $$ \times $$ 4200 J/kg/K $$ \times $$ (60 - 10) = 2 kg $$ \times $$ 4200 J/kg/K $$ \times $$ 50 K = 4200 $$ \times $$ 50 $$ \times $$ 2 J = 420,000 J
Pinput = 2000 W = 2000 J/s
Pactual = 2000 $$ \times $$ 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 10°C to 60°C is approximately 300 s.
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