Room temperature: 19 C
Temperature of water (before): 18 C
Temperature of water (after): 19 C
Oven temperature: 94 C
Mass of dry calorimeter: 80.60g=0.08kg
Mass of 100mL if water in the calorimeter: 179.18g=0.18kg
Mass of 100mL of water: 98.6g=0.0986kg

These are the results I got from the experiment. I am not sure how to determine the specific heat. There are like 10 questions for this experiment and these are the one that I couldn’t figure out. I tried calculating the specific heat with the formula mcT=mcT, but just didn’t get it. Showing the whole working would be nice. Thx :D

1 Using the specific heat of water (4180J kg-1 K-1) calculate the amount of heat absorbed by water.

2. Using the specific heat of copper (390J kg-1 K-1) calculate the amount of heat absorbed by copper.

3. The total head absorbed (heat absorbed by copper plus heat absorbed by water) is the heat lost by the brick. Using this value, calculate the specific heat of the brick.

4. Does the brick have a high, low or moderate specific heat? Why?

Quote:

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Room temperature: 19 C
Temperature of water (before): 18 C
Temperature of water (after): 19 C
Oven temperature: 94 C
Mass of dry calorimeter: 80.60g=0.08kg
Mass of 100mL of water in the calorimeter: 179.18g=0.18kg
Mass of 100mL of water: 98.6g=0.0986kg

These are the results I got from the experiment. I am not sure how to determine the specific heat. There are like 10 questions for this experiment and these are the one that I couldn't figure out. I tried calculating the specific heat with the formula mcT=mcT, but just didn't get it.

1 Using the specific heat of water (4180J kg-1 K-1) calculate the amount of heat absorbed by water.

2. Using the specific heat of copper (390J kg-1 K-1) calculate the amount of heat absorbed by copper.

3. The total head absorbed (heat absorbed by copper plus heat absorbed by water) is the heat lost by the brick. Using this value, calculate the specific heat of the brick.

4. Does the brick have a high, low or moderate specific heat? Why?

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#1 You can calculate the ΔT for water (1 C) and the mass of water (although from your information, it appears that you have two different values for the weight of water: 0.18 Kg and 0.0986 Kg).



You were given so you can calculate ΔH, the answer to #1.

As for #2, #3 and #4, you have not given enough information on the brick or copper to do the calculation.

The mass of 100mL of water is 0.0986kg. The other one is just the measurement of the water+the calorimeter.

Can you show me how to work out the specific heat for question 1?

Quote:

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Originally Posted by whatevaxd

the mass of 100mL of water is 0.0986kg. The other one is just the measurement of the water+the calorimeter.

can you show me how to work out the specific heat for question 1?

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The specific heat of water is already given in question 1. You are asked to give the amount of heat absorbed. Use the formula that Perito gave you. You perhaps got confused, I don't know. I'll put it on simpler terms.



Q is the heat involved, m the mass of water, c the specific heat capacity of water and delta theta the temperature change.

Delta theta is 1 Kelvin (from 19 - 18 Celcius)
m, you just figured it out to be 0.0986 kg
and c is given as 4180 J/ (kg K)

So,

Also, I am really confused about the other questions. Where did they mention copper or brick in the data given? It is as if these came from nowhere :confused:

Yea.. about the copper.. I have no idea as well.. it was on the investigation sheet. I am as confused as you are

Ok, the only think I can guess is that the calorimeter is made of copper. At this point, can you post the entire question? We need info about how the experiment is carried out and the mass of the brick

Yea I think so.. the calorimeter had a copper color. Before this experiment, the teacher prepared the brick for us. He put the brick with the calorimeter inside the oven. The temperature was at 94 Celsius.

I thought about it.
is the value I put in the formula correct?
Q=0.08kg x 390 x (25-94)=-2152.8J
the energy was released by 2152.8 J when it is taken out from the oven?

for this question
3. The total head absorbed (heat absorbed by copper plus heat absorbed by water) is the heat lost by the brick. Using this value, calculate the specific heat of the brick.

I am not sure how to calculate it.

I did the experiment and when it was at around 40 seconds, the temperature of the brick( while it was in the water) kept at 25 Celsius and stayed the same temperature for about 10 minutes. Does this mean that the brick have high specific heat? Or is it low?

Remember that the specific heat has to do with heat capacity per unit weight. So, if the mass of the brick is large, it might be able to absorb a lot of heat even though the specific heat isn't all that great.

If something has a low specific heat, a given quantity of it will show a rapid temperature rise when heat is added. If it has a high specific heat, it requires more heat input to heat it up.

The brick did take a long time to heat up. It also takes quite awhile for it to coll back down. Does this mean that it has high specific heat?

Not necessarily. It means it has a high heat capacity. Dividing the heat capacity by the mass gives the specific heat.

how do we do that? Heat capacity =Q?

Q is the amount of heat
c is the specific heat capacity
C is the heat capacity.

The formula for heat capacity is



And the units of C are J/K

To find the specific heat capacity, divide the heat capacity by the mass of the brick.