# Specific heat of water

1. Nov 1, 2012

### Litcyb

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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, isnt 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...

2. Nov 1, 2012

### Staff: Mentor

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

3. Nov 1, 2012

### 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