Question

Question #72765

Answer 1

`2.2 "J"/("g" ""^@"C")`

Explanation:

The problem wants you to use the given information to find the liquid's specific heat, sometimes called the specific heat capacity, of the liquid.

As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

So, you know that the liquid has a mass of `"10 g"`. Moreover, adding `"60 J"` of heat to this sample will increase its temperature by

`DeltaT = T_"final" - T_"initial"`

`DeltaT = 12.9^@"C" - 10.2^@"C" = 2.7^@"C"`

The equation that establishes a relationship between the heat needed, `q`, the mass of the sample, `m`, the specific heat of the substance, `c`, and the change in temperature, `DeltaT`, looks like this

`color(blue)(q = m * c * DeltaT)`

Rearrange this equation to solve for `c`, the specific heat of the substance

`q = m * c * DeltaT implies c = q/(m * DeltaT)`

Plug in your values to get

`c = "60 J"/("10 g" * 2.7^@"C") = 2.22"J"/("g" ""^@"C")`

You should round this off to one significant figure, since that's how many sig figs you have for the amount of heat added and for the mass of the liquid, but I'll leave it rounded to two sig figs, just for good measure

`c = color(green)(2.2"J"/("g" ""^@"C")`

So, what does this result tell you? In order to increase the temperature of `"1 g"` of this liquid by `1^@"C"`, you need to provide it with `"2.2 J"`.