Question

In an experiment to determine the rate at which heat is extracted from a hot material, 317 g of the material at 90 °C is put into a 50 mm diameter copper vessel to a depth of 90 mm as shown below. The hot material and calorimeter are placed in a container holding 221 g of cold water at 22.7 °C. The temperature of the water is then measured every 30 seconds for 3minutes. From the data provided and the graph below, determine the average heat power ( in kW ) transferred to the water

Answer 1

graph please ??

Answer 2

From the question statement and from the figure, it would seem that the only real function of the copper vessel is to serve as a container for the material to effectively transmit heat, copper being a good conductor.

Since the question asks for the AVERAGE rate, we need only the initial and final values of temperatures from the graph. Initially the temperature was 22.7 and after 3 minutes it was 24.2. So, in 180 s, the 221 g of cold water heated up by 2.5 degrees due to the heat lost by the material. Since specific heat of water is 4.2 J/gK, we can see that the heat lost by the material was 221*4.2*2.5 J (mass*sp heat*temperature change). It took three minutes to lose so much heat energy. If we divide it with this time, we get the average rate of transfer of heat.

Hopefully, that answers your question.

Answer 3

@Aditya790 , we shouldn"t take into account the the change in temperature of hot material initially at 90 oC. ?? That's the only thing that puzzled me in the morning today . Rest of your solution matches with mine.

Answer 4

I tried appying Q = mc (delta)T for the inner material at 90 oC and that's where I really got confused. Any clarifications ??

Answer 5

oh so it was Q=cm(Tf-Ti) formula...then you divide by 180 seconds...I dunno why but the way the question was asked confused me with the p=Q/t=kA((Th-Tc))/L formula...

Answer 6

@shivam_bhalla
The graph doesn't tell you about the temperature of the material, just of the water. In a real experiment done that way, it is easier to measure temperature of water. So they let the cold water heat up and measure the temperature change.

The water did heat up by 2.5 degrees. That heat had to come from the material. Since we know how much water is present (221g), we know the heat that is required for the heating. Assuming no heat dissipation, this heat energy comes solely from the material.

When you tried applying q=mcdT for the material, you didn't know what c was and there is no way to know it from the data alone.There is another thing as well, we don't know if after the three minutes, whether the water and material have reached equilibrium or not. So we don't even know the final temperature.

Starting the question from the water works because we know what happens to the water with time, how its temperature changes and its specific heat (not given, but assumed to be known in most questions). We looks at what happens to the water and using that data, we can try to predict what happens to the material.

It is like investigating the scene of a crime.You can't know what happened since you weren't there. Still, you can look around, find clues like that blood stain here or a fingerprint there or perhaps a broken vase near the bed and predict what happened. Similarly, you can't know what happened to the material as it is submerged in water. You can know, however, everything you need to about the water and then deduce.

Answer 7

Thank you for the explanation. That helps alot as i was looking for the specific heat of copper before when i tried that formula and assumed i was using the wrong formula...