When 27.2 g of solid A, at its melting point of 34.0 degree C, was added to 62.7

Question

When 27.2 g of solid A, at its melting point of 34.0 degree C, was added to 62.7 g of liquid A at 51.2 degree C, the mixture cooled to 35.0 degree C, with no solid remaining. Given the specific heat of liquid A (s = 0.798 J/g middot degree 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 q_1 + q_2 + q_3 = 0. 34.0 degree C rightarrow 34.0 degree C rightarrow 35.0 degree C leftarrow 51.2 degree C

Explanation / Answer

1. Solid A weight = 27.2gm. at 34°C

2.Liquid A weight = 62.7gm. at 51.2°C

3. Reached temperature = 34°C and phase is liquid

4.The excess heafter remaining after consumption for temperature equalization for the given masses of Solid A and Sliquid A is used to Heat of fusion for solid A

5.Heacapacity liquid A = 0.798 J/C gm

6. Heat required solid A from 34°C to 35°C for 27.2 gm is 21.70

7. Heat released from 62.7gm of liquid A to reach the temp 35°C from 51.2°C is 868.73 J

8. So the consumed for fusion is 868.73 - 21.70 = 847.03J this is for 27.2 gm so for 1gm it is equal to = 847.03/27.2=31.14J

9.So ,heat of fusion for solid A = 31.14 J/°C gm

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