IIT JEE , AIEEE Forum - Physics , Thermal Physics

coolraghu

2 kg of ice at -20 degree celcius is mixed with with 5 kg of water at 20 degree celcius ,find the water content of the final mixture

johri_anshuman

Amount of heat that can be supplied by 5 kg water at 20⁰C=5*1*20=100 Kcal

Amount of heat absorbed by ice to reach 0^oC from -20^oC=2*0.5*20=20 Kcal

remaining heat=80 Kcal

so,

80=mL=>m*80=80

=>m=1 kg

So final water content=5kg+1kg=6kg

Pls reply if i m correct.

abhijeet_0201

in the first line it should be 100kcal

edison

Well, the final content of the mixture = 5 + 2 = 7kg of water if we consider a closed system.
However, the temperature of the final mixture is different and is given by taking care of the fact that specific heat of ice = 0.5Cal/g/deg C
and specific heat of water = 1cal/g/ deg C

Also take care of latent heat of ice from changing the state to liquid.

Use this and obtain the final result for the temperature of the mixture.

Ask & Discuss Questions with Community & Experts