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# Water of volume 2 litre in a container is heated with a coil of 1kW at 27°C. The lid of the container is open and energy dissipates at the rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C? [Given specific heat of water=4.2kJ/kg]

Md asharuddin
11 Points
4 years ago
The heater coil gives energy at a rate of 1000 J/s, out of which 160 J/s is dissipated through the lid. Thus, energy for heating the water is available at a rate of,P=1000−160=840J/s.P=1000−160=840J/s.The heat energy required to raise the temperature of mass mm of water from T1T1 to T2T2 is,Q=mS(T2−T1),Q=mS(T2−T1),where SS is specific heat of the water. If tt is the time required to raise temperature of mass mm from T1T1 to T2T2 then, by energy conservation,Pt=Q.Pt=Q.The density of water is 1000kg/m31000kg/m3. Thus, mass of 2 litre of water is m=2kgm=2kg. Substitute PP and QQ from first and second equations into third equation to get,t=mS(T2−T1)P=(2)(4.2×103)(77−27)840=500second=8min20second.
Yash Chourasiya
one year ago
Dear Student

Please see the solution in the attachment.