A 28 g block of ice is cooled to −78◦C. It is added to 562 g of water in an 80 g copper calorimeter at a temperature of 21◦C.
Find the final temperature. The specific heat of copper is 387 J/kg ·◦C and of ice is 2090 J/kg ·◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg ·◦C .
Answer in units of ◦C.
i need help asap
Mass of ice = mi = 28 g = 0.028 kg
Mass of water = mw = 562 g = 0.562 kg
Mass of the copper calorimeter = mc = 80 g = 0.08 kg
Specific heat of copper = Cc = 387 J/(kg.oC)
Specific heat of water = Cw = 4186 J/(kg.oC)
Specific heat of ice = Ci = 2090 J/(kg.oC)
Latent heat of fusion of water = L = 3.33 x 105 J/kg
Initial temperature of ice = T1 = -78 oC
Melting point of ice = T2 = 0 oC
Initial temperature of water and the copper calorimeter = T3 = 21 oC
Final temperature of the mixture = T4
The heat gained by the ice is equal to the heat lost by the water and the copper calorimeter.
miCi(T2 - T1) + miL + miCw(T4 - T2) = mwCw(T3 - T4) + mcCc(T3 - T4)
(0.028)(2090)(0 - (-78)) + (0.028)(3.33x105) + (0.028)(4186)(T4 - 0) = (0.562)(4186)(21 - T4) + (0.08)(387)(21 - T4)
4564.56 + 9324 + 117.208T4 = 49403.172 - 2352.532T4 + 650.16 - 30.96T4
2500.7T4 = 36164.772
T4 = 14.46 oC
Final temperature = 14.46 oC
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