Water flows at the rate of `0.1500 kg//min` through a tube and is heated by a heater dissipating `25.2W`. The inflow and outflow water temperatures are `15.2^@C` and `17.4^@C`, respectively. When the rate of flow is increased to `0.2318 kg//min` and the rate of heating to `37.8W`, the inflow and outflow temperature are unaltered. Find
i. the specific heat capacity of water
ii. the rate of loss of heat from the tube.

Explanation

i. `(Deltam)/(Deltat)=(0.15)/(60)kg//s=2.5xx10^-3(kg)/(s)`
Let P be the rate of loss of heat from the tube, and C be the specific heat capacity of water. Then
`P+DeltamxxCxx(17.4-15.2)=25.2` ..(i)
Also , `P+Delta' mxxCxx(17.4-15.2)=37.8` ..(ii)
where `Delta'm=(0.2318)/(60)=3.8633xx10^-3kg//s`
Eq. (ii)- Eq. (i)
`2.2xxC[3.86-2.5]xx10^-3=12.6`
`C=4.2xx10^3(J)/(kg^@C)`
ii. Putting the value of C in Eq. (i) `P+23.1=25.2`
`P=2.1W`