Question

A 0.033 kg glass (with c = 840 J/kg oC) contains 0.281 of lemonade which, due to the sugar content, has a specific heat of 4,208 J/kg oC. After putting 0.049 kg of ice into the glass and allowing it to completely melt the final equilibrium temperature of the glass of lemonade is found to be 2.9 oC. intial temperature is 0.

(a) Calculate the initial temperature of the lemonade and glass. oC

Note the following data for ice/water: specific heat (ice)= 2100 J/(kg*oC) , Latent heat of fusion = 3.33 x 105 J/kg , melting/freezing point = 0 oC, specific heat (water) = 4186 J/(kg*oC)

Answer #1

Let the initial temperature of lemonade and glass be T degree C

Heat lost by lemonade and glass is

Heat gained by ice is

Equating the heat lost to heat gained

Solving this gives the initial temperature of the lemonade and glass

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