Question

Question #ce46f

Answer 1

`33.3^@"C"`

Explanation:

The key to this problem is the specific heat of olive oil, which is said to be equal to

`c_"olive oil" = "2.19 cal g"^(-1)""^@"C"^(-1)`

As you know, the specific heat of a substance tells you how much heat must be added to `"1 g"` of that substance in order to increase its temperature by `1^@"C"`.

In your case, you must add `"2.19 cal"` of heat for every `"1 g"` of olive oil in order to increase its temperature by `1^@"C"`.

Now, the first thing to do here is to figure out how much heat is needed in order to increase the temperature of `"55.0 g"` of oil by `1^@"C"`

`55.0 color(red)(cancel(color(black)("g"))) * "2.19 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "120.45 cal"^@"C"^(-1)`

This means that in order to increase the temperature of `"55.0 g"` of olive oil by `1^@"C"`, you need to supply it with `"120.45 cal"` of heat.

Consequently, adding `"877 cal"` of heat will produce an increase in temperature of

`877 color(red)(cancel(color(black)("cal"))) * overbrace( (1^@"C")/(120.45color(red)(cancel(color(black)("cal")))))^(color(blue)("for 55.0g of olive oil")) = 7.28^@"C"`

Therefore, the final temperature of the olive oil will be

`color(darkgreen)(ul(color(black)(T_"final" = 26.0^@"C" + 7.28^@"C" = 33.3^@"C")))`

The answer is rounded to three sig figs.