Question

an electric heater rated at 2 kilowatt is used to hate 200 kg of water from 10 degree Celsius to 70 degree Celsius assuming no heat losses the time taken East options are 2.2 5.2 seconds 6 into 10 raise to power 3 seconds 25.2 into 10 raise to power 3 seconds 25.2 into 10 raise to power 3 seconds​

Answer 1

Hey bf ,

Here is solution !

Given, P = 2 kW = 2000 W

​Mass of water = 20 kg = 20,000 g

Rise in temperature, dθ = θ2 - θ1 = 30 - 10 = 200C

​We know that specific heat of water, (c)= 1 cal g-1 0C​-1

Heat produced by heater in time t = P X t = 2000 X t joule = 2000 t4.2 calories (since, 1 cal = 4.2 Joules)

Heat taken by water = mc ​ (θ2 - θ1 ) = 20,000 X 1 X 20= 4,00,000 calories

As, heat produced by heater = Heat taken by water

Hence, 2000 t4.2 = 4000000n solving, we get :t = 840 s

@Badshah