Properties Of Solids And Liquids Question 109
Question: Water of volume 2 litre in a container is heated with a coil of $ 1kW $ at $ 27{}^\circ C $ . The lid of the container is open and energy dissipates at rate of $ 160J/s. $ In how much time temperature will rise from $ 27{}^\circ C $ to $ 77{}^\circ C $ [Given specific heat of water is $ 4.2kJ/kg $ ] [IIT-JEE (Screening) 2004]
Options:
A) 8 min 20 s
B) 6 min 2 s
C) 7 min
D) 14 min
Show Answer
Answer:
Correct Answer: A
Solution:
Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to environment)
therefore $ mc\ \Delta \theta =P _{Coil}\ t-P _{Loss}\ t $
therefore $ 2\times 4.2\times 10^{3}\times (77-27)=1000t-160\ t $
therefore $ t=\frac{4.2\times 10^{5}}{840}=500\ sec\ =8\ min\ 20\ sec $
