Question

Question #c1015

Answer 1

`50^@"C"`

Explanation:

In order to solve this problem, you need to know water's specific heat, which expresses the amount of heat needed to raise the temperature of `"1 g"` of water by `1^@"C"`.

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

Now, the amount of heat is given to you in calories, which means that you're going to have to convert it to Joules

`1000color(red)(cancel(color(black)("cal"))) * "4.18400 J"/(1color(red)(cancel(color(black)("cal")))) = "4184.0 J"`

The equation that establishes a relationship between heat added/removed and increase/decrease in temperature looks like this

`q = m * c * DeltaT" "`, where

`q` - heat absorbed/removed;
`m` - the mass of the substance;
`c` - its specific heat;
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature.

Rearrange the equation to solve for `DeltaT`

`DeltaT = q/(m * c)`

`DeltaT = (4184.0color(red)(cancel(color(black)("J"))))/(100color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ^@"C")) = 10.0 ""^@"C"`

This means that the final temperature of the water will be

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

`T_"final" = "DeltaT" + T_"initial" = 10""^@"C" + 40""^@"C" = color(green)(50""^@"C")`