A beaker contains 40 g of water at 20 degree C. Now 50 g of ice is put into the beaker. The resulting temperature will be
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this problem is done assuming that all ice melts to mix with water
first u need to know specific heat of water=4200J/kgK
The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.
heat gained by water=heat lost by ice
heat gained by water=ms(T2=T1)
s is specific heat
heat lost by ice=Q=mL+ms(T2-T2')
T2'=o degrees(starting temp of moten water)
where l is latent heat of ice(heat req to change it from solid to liquid sate)
we have heat req to change the stae of ice+heat req to raise temp of melted water(as final temp is not 0degrees celcius)
L=335J/g (latent heat of water)
equate both to find T2