1 gm ice at -40 deg Celsius is placed in a container havng 1gm water at 1 deg Celsius.  Find equlilbrium temperature

1 gm ice at -40 deg Celsius is placed in a container havng 1gm water at 1 deg Celsius.  Find equlilbrium temperature

heat reqd to raise the temp of ice to 0 degree C = 1*0.5*40 = 20 cal

heat reqd to conver ice at 0 degree C to water at 0 degree C = mL = 1*80 = 80 cal

so total = 100 cal

But to conver 1 gm of water at 1 degree C to 1 gm of ice at 0 degree C , heat reqd = 1+ 80 = 81

so temp of mixture = 0 degree C

Heat required to cool water to level of ice = mCx1

where C is specific heat of water

Heat required to convert water into ice = mL

where L is latent heat of fusion

Heat required to cool ice from water = mC' x t

where t is final temp of mixture

Heat required to raise temp of ice = m C' x (40-t)

where C' is specific heat of ice

mC + mL + mC't = mC'(40-t)

m = 1 so it cancels

place values of C and C' and calculate t

no that won't work here.

see even if 1 gm of water at 1 degree C is freezed , it will liberate only 81 cal.

Whereas to convert the given ice to water 100 cal is needed.

so the temp of the mixture will be ) degree C