A cup of newly brewed 330 g cofee is to hot to drink when served at 88 degree ce

Question

A cup of newly brewed 330 g cofee is to hot to drink when served at 88 degree celcisus.

a)IF you add 60 g of ice at -15 degree celsisus to it. What is the final termpetate Tf of the cofee when it reaches thermal equilibrium?

b)What is the mass of the ice cube needed in order to cool the coffee to a pleasant 50 degree celsisus? assusming specific heat of coffee is 4190 J/kg * C and the specific heat of ice is 2050 J/kg * C. The latent heat of fusion is Lf= 334 x 10^3 J/kg and the latent heat of vaporization of water is Lv= 2.26 x 10^6 J/kg

Explanation / Answer

(a) At thermal equilibrum Tf is the tempratue of mixture

Mass of coffee = 330 g = 0.33 kg

Mass of ice= 60 g = 0.06 Kg

Heat lost by coffee = Heat gained by ice

First ice get heated to 0 celsius

Then ice melts

Then water from melted ice is heated to Tf

m1*s1*dT1= m2*s2*dT2 + m2*Lf + m2*s1*dT3

0.33*4190*(88-Tf) = 0.06( 2050*(0+15) + 334000 + 4190*(Tf - 0) )

Solve ang get Tf = 61.06 Celsius

(b) Use the equation again And put Tf = 50

m1*s1*dT1= m2*s2*dT2 + m2*Lf + m2*s1*dT3

0.33*4190*(88-50) = m( 2050*(0+15) + 334000 + 4190*(50 - 0) )

Solve and get m = 0.0915 Kg = 91.5 g

So 91.5 gm ice is needed

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