Specific Heat of Water - What is the Heat Capacity?

[Litcyb]

Homework Statement

What is the heat capacity of water in J*g*°C

Homework Equations

A 74.8 sample of copper at 143.2g
is added to an insulated vessel containing 165ml of water, Density of water = 1.00 at 25.0 °C.
The final temperature is 29.7° C
The specific heat of copper is 0.385 J*g*°C.

The Attempt at a Solution

g of water =165g
quantity of heat= mass of substance * specific heat* temperature change
heat capacity = C (mass of substance * specific heat)

quantity of heat of copper = 74.5g*0.3858(29.7-143) =-3.268*10^-3
so, isn't this the same quantity of heat of water? if so, this how i tried to do it

165g of h2o* specific heat*(29.7-25)= 3.268*10^-3

specific heat =4.2147
so, 165*4.21= 700...
but that's wrong, the right answer is 74.5g HELP PLEASE!

 

[Borek]

How come right answer is 74.5 g if the question is about specific heat capacity (which you calculated earlier)?

 

[Litcyb]

YES! So sorry, I wrote that all wrong and didn't know how to fix it! my units weren't correct.

It took me a while to see my errors but i managed to solve it. In reality, they were asking for J*mol*°C
In order to calculate that we just simple do the following
the total quantity heat of copper which is 74.8*0.385*(143.2-29.7)
q of copper = -3268.57 -->-3.26*10^-3
so like no energy can be destroyed or created, the total quantity of heat for water is the opposite which is 3.26*10^-3
with that being said, now we could calculate the heat capacity and then turn them into moles.
which is done by the following,
3268.7 J*g*°C = 165g of H2O * specific heat* (29.7°C-25°C)
3268.7J*g*°C= Specific heat *775.50

Specific heat of water = 4.21 J*g*°C , NOW i need to convert that into moles of by multiplying by the molar mass of H2O (18g).

Final answer is 4.21 * 18 = 75.85 ≈75.9 J*mol*°C

Thanks for answering anyways!

 

[epenguin]

Mynah myah that took me back to school, especially the number which I remembered, let's see that was 4.1 something and then I see 4.12 and it came back. Because that is what used to be called The Mechanical Equivalent of Heat and memory is of countless boring exercises on it converting calories into joules etc. (I think we may even done some of them with non-metric units like food-pounds as well).

There seemed to be even an ideological aura attached to it because it was about Work not fun or other vanities and came from the no-nonsense working industrial city of Manchester.

I had not realized till today that Joule's work had generated so much heat: http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat

 

[DeeNos]

The heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. In the case of water, the heat capacity is 4.184 J/g°C. This means that it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

In the given scenario, the heat capacity of water can be calculated by using the formula:
heat capacity = mass of water * specific heat * temperature change

Substituting the given values, we get:
heat capacity = 165 g * 4.184 J/g°C * (29.7°C - 25°C) = 2,764.22 J/°C

This means that it takes 2,764.22 joules of energy to raise the temperature of 165 grams of water by 4.7 degrees Celsius.

To calculate the final temperature of the water after adding the copper, we can use the law of conservation of energy, which states that the amount of heat lost by the copper must be equal to the amount of heat gained by the water.

Therefore, we can set up the equation:
heat lost by copper = heat gained by water
mass of copper * specific heat of copper * (final temperature - initial temperature) = mass of water * specific heat of water * (final temperature - initial temperature)

Substituting the given values, we get:
74.8 g * 0.385 J/g°C * (final temperature - 143.2°C) = 165 g * 4.184 J/g°C * (final temperature - 25°C)

Solving for the final temperature, we get:
final temperature = (74.8 * 0.385 * 143.2 + 165 * 4.184 * 25) / (74.8 * 0.385 + 165 * 4.184) = 29.86°C

Therefore, the final temperature of the water is 29.86°C. This is slightly different from the given value of 29.7°C, which could be due to rounding errors or slight variations in the specific heat values used.

In conclusion, the heat capacity of water is 4.184 J/g°C, and it takes 2,764.22 joules of energy to raise the temperature of 165 grams