Question

A 100 Ω resistive heater in a tank of water (1kg) increases its temperature from 10°C to 20°C over a period of 1 hour. During this period of time your measurements show that the voltage of the heater was 50V. How much energy is not absorbed by the water and leaves the system into the srroundings? Assume constant specific heat and density.

Answer 1

Answer:48.52 kJ

Step-by-step explanation:

Given

Resistance
`=100 \Omega`

temperature increases from
`10^(\circ)C to 20^(\circ)C`

Voltage=50 V

Heat given(H)
`=(V^2t)/(R)`

Where V=voltage applied

t=time

R=Resistance

`H=(50^2* 60* 60)/(100)=90 kJ`

Heat absorbed by water is

`Q=mc(\Delta T)`

where

m=mass of water

c=specific heat of water

`\Delta T`=change in temperature

`Q=1000* 4.184* (20-10)=41.48 kJ`

Therefore 90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.