A cube of solid benzene (C6H6) at its melting point and weighing 6.52 g is placed in 21.8 g of water at 39.0°C?
Given that the heat of fusion of benzene is 9.92 kJ/mol, to what temperature will the water have cooled by the time all of the benzene has melted. You can assume that the benzene remains at its melting point.
More Questions From This User See All
Answers & Comments
Verified answer
You also need to know the specific heat of liquid water which is 4.18*J/g-C and the mass of benzene per mole which is 78.11g/mol.
For all the benzene to melt, calculate the heat:
Q = H*m = 6.52g*9.92kJ/mol*(1mol/78.11g) = .8280425040*kJ
Now find the change in temperature for water, where Td is temperature difference and Cv heat capacity:
Q = Cv*Td*m
solving for Td:
Td = Q/(Cv*m) = (.8280425040*kJ)/(4.18*J/g-C*21.8g) = 9.09C
So subtract that temperature difference (because it cooled) to the starting temparture:
T = T - Td = 39.0C - 9.09C = 29.91C
T = 29.91