Water, flowing through a pipe at the rate of `"0.15 kg.min"^(-1)`, is heated by a 25.2 W heater. Temperature of the incoming and the outgoing water are `15.2^(@)C` and `17.4^(@)C` respectively. When the rate of flow of water is increased to `"0.2318 kg.min"^(-1)` and rate of heating to 37.8W, temperature of incoming and outgoing water remain the same. Find the specific heat capacity of water and the rate of loss of heat through the pipe.

Explanation

Let specific heat of water be `"s.J.kg"^(-1).^(@)C^(-1)` and rate of heat loss be `"H J .s"^(-1)`.
From the given data,
`=(0.15)/(60)xxsxx(17.4-15.2)+H=25.2" …(1)"`
`and (0.2318)/(60)xxsxx(17.4-15.2)+H=37.8" ...(2)"`
Subtracting equation (1) from equation (2) we get,
`(sxx22)/(60)(0.2318-0.15)=12.6`
`"or, "s="4200 J .kg"^(-1).^(@)C^(-1)`
Then from equation (1),
`H=25.2-(0.15)/(60)xx4200xx2.2`
`=2.1"J.s"^(-1)=(2.1)/(4.2)"cal.s"^(-1)`
`="0.5cal.s"^(-1)`