How many grams of steam at 100 degree C would be required to raise the temperatu

Question

How many grams of steam at 100 degree C would be required to raise the temperature of 35.8 g solid benzene (C_6 H_6) from 5.5 degree C to 45.0 degree C? Assume that heat is only transferred from the steam (and not liquid water) and that the steam/water and benzene are separated by a glass wall and do not mix. (The melting point of benzene is 5.5 degree C; Delta H_fux for benzene is 9.87 kJ/mol; specific heat for benzene is 1.63 J/g middot degree C; Delta H_vap for steam at 100 degree C is 40.7 kJ/mol.)

Explanation / Answer

Heat required for 35.8 g solid benzene to convert liquid benzene at 5.5 deg C=m*latent heat of melting=38.5*9.87 kJ

Now to increase temperature of 35.8 g liquid benzene from 5.5 deg C to 45.0 deg C=m*s*t where m=mass, s=specific heat and t=temperature rise

So heat required=38.5*1.63*39.5/1000 kJ=2.478 kJ

So total heat required=382.473 kJ

Now 1 g steam can release 537*4.184/1000 kJ heat. So steam of 100 Dec C is required=382.473/2.246 g=170.29 g

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