one kilogram of ice is droppedd in 9 kg of water at 50 degree celcius. What will be its final temperature while the latent specific heat of fusion of ice is 336000 J/Kg?
the answer is 37 degree celcius or 310 K......i applied many method but in vain.
Here we need to understand the scenario. When the ice cube dropped in to the water initially we have solid ice at 0C and 50C of water.Then due to the heat of water ice cube will dissolve and increase its temperature.
In the first stage latent heat of fusion will need to make 0C solid ice to water at 0C. This is given by the 50C water. Then the heat of the ice cube (which now is water) will increase its temperature while 50C water will decrease. When both come to a same temperature the heat exchange will stop and the system will come to equilibrium.
P = Energy required to make 0C ice to 0C water
P = latent heat of fusion*mass
`P = 336000xx1`
`P = 336000J`
Let us say the final temperature of the system is 'T'.
We consider specific heat capacity of water (C) as 4186J/(kg*C)
Then using law of conservation of energy at equilibrium;
Heat loss by 50C water = Heat absorbed by ice
`m_1xxCxxDeltaT = P+m_2xxCxxDeltaT`
`9xx4186xx(50-T) = P+1xx4186xx(T-0)`
`Txx4186(9+1) = 9xx4186xx50-336000`
`T = 36.97`
So the final temperature of the system is 36.97C or 37C.