Question

The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

Answer 1

`"239 J"`

Explanation:

First thing first, you mistyped the specific heat of water, which should be

`c_"water" = 4.18"J"/("g" ""^@"C")`

Now, a substance's specific heat tells you how much heat is required to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In the case of water, you would need `"4.18 J"` to increase the temperature of `"1 g"` of water by `1^@"C"`.

Notice that your sample of water has a mass of `"1 g"` as well, which means that the only factor that will determine the amount of heat needed will be the difference in temperature.

The equation that establishes a relationshop between heat and change in temperature looks like this

`color(blue)(q = m * c * DeltaT)" "`, where

`q` - heat absorbed
`c` - the specific heat of the substance, in your case of water
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values and solve for `q` to get

`q = 1.00color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (83.7 - 26.5)color(red)(cancel(color(black)(""^@"C")))`

`q = "239.096 J"`

Rounded to three sig figs, the answer will be

`q = color(green)("239 J")`

SIDE NOTE Notice how much heat would be needed to increase the temperature of 57.2 g of water by `1^@"C"`. `DeltaT` will now be equal to `1^@"C"` and you'd have

`q = 57.2color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 1color(red)(cancel(color(black)(""^@"C")))`

`q = "239 J"`

This tells you that you need the same amount of heat to increase the temperature of `"1 g"` of water by `57.2^@"C"`, as you do to increase the temperature of `"57.2 g"` of water by `1^@"C"`.