(A) The heat required to raise the temperature of the water is given by $Q = mc \Delta \theta$.Here,mass $m = 2 \, kg$ (since density of water is $1 \

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(A) The heat required to raise the temperature of the water is given by $Q = mc \Delta \theta$.Here,mass $m = 2 \, kg$ (since density of water is $1 \, kg/L$),$c = 4.2 \times 10^3 \, J/(kg \cdot K)$,and $\Delta \theta = 77 - 27 = 50 \, ^\circ C$.So,$Q = 2 \times 4.2 \times 10^3 \times 50 = 4.2 \times 10^5 \, J$.The net power supplied to the water is $P_{net} = P_{coil} - P_{loss} = 1000 \, W - 160 \, W = 840 \, W$.The time taken $t$ is given by $t = Q / P_{net}$.$t = (4.2 \times 10^5) / 840 = 500 \, s$.Converting to minutes: $500 \, s = 8 \, min \, 20 \, s$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

Medium

Water of volume $2 \, L$ in a container is heated with a coil of $1 \, kW$ at $27 \, ^\circ C$. The lid of the container is open and energy dissipates at a rate of $160 \, J/s$. In how much time will the temperature rise from $27 \, ^\circ C$ to $77 \, ^\circ C$? [Given: specific heat of water is $4.2 \, kJ/(kg \cdot K)$]

IIT 2004 All IIT

A
$8 \, min \, 20 \, s$
B
$6 \, min \, 2 \, s$
C
$7 \, min$
D
$14 \, min$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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