Question

What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?

Answer 1

`66^@"C"`

Explanation:

In order to be able to solve this problem, you need to know the value of water's specific heat, which is usually given to be `4.18"J"/("g" ""^@"C")`.

A substance's specific heat tells you how much heat is required to increase the temperature of `"1.0 g"` of that substance by `1^@"C"`.

Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this

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

`q` - the amount of heat
`m` - the mass of the sample
`c` - its specific heat
`DeltaT` - the change in temperature, defined s final temperature minus initial temperature

So, in order to find the final temperature of the water sample, you need to first find the change in temperature, `DeltaT`. Since heat was absorbed by the water, you must have a positive value for `DeltaT`.

Plug in your values into the above equation to get

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

`DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"`

Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that

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

`T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")`

The answer is rounded to two sig figs, the number of sig figs you have for the initial temperature and the mass of the sample.