Water of volume 2 litre in a container is heated with a coil of 1,kW at 27,° C . lid of container i

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Water of volume $2$ litre 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 rate of $160\,J/s.$ In how much time temperature will rise from $27\,^\circ C$ to $77\,^\circ C$ [Given specific heat of water is $4.2\,kJ/kg$]

A

$8 \,min \,20 \,s$

B

$6 \,min \,2 \,s$

C

$7\, min$

D

$14 \,min$

(IIT-2004)

Solution

(a) Heat gained by the water $=$ (Heat supplied by the coil) $-$ (Heat dissipated to environment)

==> $mc\;\Delta \theta = {P_{Coil}}\;t – {P_{Loss}}\;t$

==> $2 \times 4.2 \times {10^3} \times (77 – 27) = 1000\,t – 160\;t$

==> $t = \frac{{4.2 \times {{10}^5}}}{{840}} = 500\;sec\; = 8\;min\;20\;sec$

Standard 11

Physics

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