Hi I need help going about this problem. I calculated the heat required using th

Question

Hi I need help going about this problem. I calculated the heat required using the delta heat of fusion for C3H8 but the heat capacity for ice and water, is this the right first step? And if delta heat of fusion is negative, do i leave it as negative? I am confused as to how to continue the problem.

Using information in Appendices B and C in the textbook, calculate the minimum number of grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 6.65kg of ice at -16.0?C to liquid water at 72.5?C.

Explanation / Answer

HEAT REQUIRED FOR CONVERTING -16oC ICE TO 72.5oC WATER

HEAT REQ = (1)Heat req to bring 16oC ice to 0oC ice + (2)Heat for phase change of 0oC ice to 0oC water + (3)Heat to bring 0oC water to 72.5 oC water

avg cp of ice = (2.05 +1.96)/2 = 2.005 kJ/kg K

latent heat of fusion of ice = 335 kJ/kg

avg cp of water = 4.187 kJ/kg K

(1) = mCp(T2-T1) = 6.65* 2.005*(0+16) = 213.332 kJ

(2)= Hlatent *m = 6.65*335 = 2227.75 kJ

(3)= mCp (T4-T3)= 6.65*4.187*(72.5-0) = 2018.66 kJ

=> Heat req. = (1)+(2)+(3) = 4459.74 kJ

Now , Heat of combustion of C3H8 = 50.03KJ/g

Grams of C3H8 req = 4459.74/50.03 = 89.14 g

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects