Please answer part A and B. B located at bottom of screen You mix m_I = 0.85 kg

Question

Please answer part A and B. B located at bottom of screen

You mix m_I = 0.85 kg of ice at T_I = -22 degree C with m_W = 4.2 kg of water at T_W = 72 degree C in an insulated container. The specific heats of ice and water are c_I = 2.10 times 10^-5 J (kg middot degree C) and c_W = 4.19 times 10^3 J(kg middot degree C), respectively, and the latent heat of fusion for water is L_f = 3.34 times 10^5 J/kg. Enter an expression for the final, equilibrium temperature of the mixture, in terms of the defined quantities. Calculate the final, equilibrium temperature of the mixture, in degrees Celsius.

Explanation / Answer

Heat lost by water is equal to heat gained by ice

mw*cW*(Tw-Tf) =mi*Ci*(0-(-22))*Lf+cw*(Tf-0)

4.2*4190*(72-Tf) =0.85*2100*22+334000+4186*Tf

17598(72-Tf)=39270+33400+4186Tf

1267056-17598Tf =72670+4186Tf

1194386=21784Tf

Tf =54.82oC or 327.828K

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