50 grams of liquid water at 10 degrees celsius is added to 4.0 kg of solid dry i

Question

50 grams of liquid water at 10 degrees celsius is added to 4.0 kg of solid dry ice (CO2) which was initially at -90 degrees celsius. The very low temperature of the dry ice combined with its significantly larger mass causes the water to cool down and eventually freeze solid. At the moment all the water freezes, what is the change in temperature (delta T) of the dry ice? Assume the dry ice does not undergo a phase change. Specific heat capacity of water is 4.19 J/g*K, latent heat of transformation for freezing water is 333 J/g and the specific heat capacity of dry ice in this temperature range is 1.25 J/g*K.

Explanation / Answer

From principle of caloriemetry

heat gained by ice = heat lost by water

mi*Si*dT1 = (mw*Sw*dT2)+(mw*Lf)+(mw*Si*dT3)

here mi is the mass of given ice

Si is the specific heat capacity of ice

dT1 is the change in temparature of ice = 90-T

mw is the mass of thewater = 0.050 kg
dT2 = 10 degrees celsius

dT3 = T degrees celsius

Then

From the above equation

4*1.25*10^3*(90-T) = (0.050*4190*10)+(0.050*333000)+(0.050*1250*T)

450000-5000T = 18745+62.5*T
(5000+62.5)*T = 450000-18745 = 431255

T = 431255/(5000+62.5) = 85.18 degres celsius

answer is -85.2 degrees celsius

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