Question

When 27.2 g of solid A, at its melting point of 34.0 °C, was added to 62.7 g of liquid A at 51.2 °C, the mixture cooled to 35.0 °C, with no solid remaining. Given the specific heat of liquid A (s = 0.798 J/g·°C), find the heat of fusion per gram of solid A. Assume that no heat enters or leaves the mixture. That is, assume that the mixture is an isolated system and therefore that q1 + q2 + q3 = 0.

Answer 1

Let the heat of fusion of solid A per gram is Hf J/g

So, to the heat required to melt 27.2g of A = mass*Hf = 27.2g*Hf J/g

= 27.2Hf J

The heat required to raise the temperature of this 27.2g of liquid A to 35 C from 34 C = mST

where m is the mass here m = 27.2g , S is the specifice heat of liquid A here S = 0.798J/g-C, T is the change in temperature of the liquid A = 35 C - 34 C = 1 C

So, heat required to raise the temperature of this 27.2g of liquid A to 35 C from 34 C =

27.2 g * 0.798 J/g-C * 1 C = 21.7 J

The amount of heat required for the above two processed must've come from the liquid A which is at higher temperature

of 51.2 C

So, the amount of heat lost by this liquid = heat gained by the 27.2 g A sample (conservation of energy in Isolated system)

the amount of heat lost by this liquid = m1ST1

m1 = 62.7 g

T1 = 51.2 C - 35 C = 16.2 C

amount of heat lost by this liquid = 62.7 g*0.798 J/g-C*16.2C

= 810.56 J

So,

Equating  the amount of heat lost by this liquid = heat gained by the 27.2 g A sample

810.56 J = 27.2Hf J + 21.7 J

Hf = (810.56-21.7)/27.2 J/g

= 29 J/g

So, the heat of fusion of solid A per gram is Hf J/g = 29 J/g