[sasha11]

When 47.5 J of heat are added to 13.2 g of a liquid, its temperature raises by 1.72 degree Celsius. What is the heat capacity of the liquid?

Q=m·c·Δt
47.5 joules = 13.2g·4.18Joules·1.72°c
47.5 = 94.90272g

Do I divide this I'm not sure if this is right?Can someone point me in the right direction?

Thanks

[UG]

What does the c stand for in your equation?

[sasha11]

c stands for specific heat

[UG]

Yes that is true, but another name for specific heat is the specific heat capacity which is what you want to find. The unknown is c, you need to rearrange your equation to isolate c and then solve using the information given.

[sasha11]

C=[47.5joules]/[13.2grams·1.72°c]

C= 2.09 j/gc

with scientific digits
I think it will be
C=2.10 j/gc
If I'm wrong tell me where please
Thanx

[UG]

If the answer comes out as 2.09 then leave as is since all other values are known to three significant figures