Answer:48.52 kJ
Explanation:
Given
Resistance[tex]=100 \Omega [/tex]
temperature increases from [tex]10^{\circ}C to 20^{\circ}C[/tex]
Voltage=50 V
Heat given(H)[tex]=\frac{V^2t}{R}[/tex]
Where V=voltage applied
t=time
R=Resistance
[tex]H=\frac{50^2\times 60\times 60}{100}=90 kJ[/tex]
Heat absorbed by water is
[tex]Q=mc(\Delta T)[/tex]
where
m=mass of water
c=specific heat of water
[tex]\Delta T[/tex]=change in temperature
[tex]Q=1000\times 4.184\times (20-10)=41.48 kJ[/tex]
Therefore 90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.
