Heat Transfer: Phase and Temperature Can someone help explain the answer very co

Question

Heat Transfer: Phase and Temperature

Can someone help explain the answer very confused on this one

The following information is given for copper at 1 atm: A 32.20 g sample of solid copper is initially at 1.054 Times 10^3 degree C. If 8.325 Times 10^3 J of heat are added to the sample at constant pressure (P = 1 atm), which of the following are true. The sample is a solid in equilibrium with liquid. The sample is a solid. The sample is a liquid. The sample is at a temperature of 1.0S3 -103 C. The sample is at a temperature greater than 1.0S3 -  103 SC. An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer.

Explanation / Answer

Solution :-

Lets calculate the amount of heat needed to change the temperature of solid from 1054 C to 1083 C

q= m*c*delta T

= 32.2 g * 0.3850 J per g C *(1083 – 1054 C)

= 359.513 J

Now lets calculate the amount of heat needed to melt the solid

q= delta H f * m

  = 204.7 J per g * 32.2 g

= 6591.34 J

Now lets calculate the amount of heat needed to raise the temperature of the liquid

q= m*c*delta T

   = 32.2 g * 0.4940 J per g C * (2567 C – 1083 C)

= 23606 J

Heat provided = 8325 J which is very less that heat needed to reach the boiling point.

So

But heat needed to melt the solid is = 359.513 J + 6591.34 J = 6950.853 J

So heat needed to melt the solid is less than heat provided

This means the 32.2 g solid sample will melt to give liquid sample and due to some excess heat available it will rise in temperature above the 1083 C

So the answers are as follows

The sample is liquid

The sample is at a temperature above 1.083*1063 C

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