Question

2. 300 g of ice at zero degree centigrade is mixed with 500 g of water at 50 degrees centigrade. What will be the final temperature of the mixture? A. 1.25 celsius B. 2.0 Celsius C. neither A nor B

Answer 1

Data we should know first :

Latent heat of fusion of ice = 334 J /g and

Specific heat capacity of water = 4.186 J / g °C

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Here , mass of the ice = 300 grams

and mass of the water @ 50 °C = 500 grams

so therefore, according to principle of Calorimetry:

Heat gained by ice = Heat lost by water

latent heat of fusion of ice + Heat gained by ice water @ 0 °C to reach at temp. T = heat lost by water from reaching a temp. T from 50 °C

=> M_i.L_i + M_w.C_w.$\Delta$T = M_w.C_w.$\Delta$T

=> (300 x 334 ) + ( 300 x 4.186 x (T_f - 0) ) = (500 x 4.186 x (50 -T_f ) )

=> 100200 + 1225.8*T_f = 104650 - 2093*T_f

=> (1225.8 + 2093 )T_f = 104650 -100200

=> T_f = 1.34 °C

so, final temperature of the mixture is T_f = 1.34 °C