Question

Question #087c4

Answer 1

`91.6^@"C"`

Explanation:

For starters, you need to make a note that the specific heat of water is equal to

`c_"water" = "1 cal J g"^(-1)""^@"C"^(-1)`

This means that when the temperature of `"1 g"` of liquid water decreases by `1^@"C"`, `"1 cal"` of heat is being given off.

Now, your tool of choice here will be this equation

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

Here

• `q` is the heat gained by the water
• `m` is the mass of the water
• `c` is the specific heat of water
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You need to solve for `DeltaT`, so rearrange the equation as

`DeltaT = q/(m * c)`

Plug in your values to find

`DeltaT = (1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = 7.2^@"C"`

Now, you know that the sample is losing heat, so you can say that the final temperature will be `7.2^@"C"` lower than the initial temperature

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

which, in your case, will be equal to

`T_"final" = 98.8^@"C" - 7.2^@"C" = color(darkgreen)(ul(color(black)(91.6^@"C")))`

The answer is rounded to one decimal place.

`color(white)(a)`

SIDE NOTE It's important to notice that if you use

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

you get

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

`color(red)(cancel(color(black)(T_"final" = 7.2^@"C" + 98.8^@"C" = 106^@"C"))) ->` not good

The problem here is that the sample is actually giving off heat, so you should use

`q = - "1794 cal"`

in the equation. The minus sign is needed because the final temperature is lower than the initial temperature. This will get you

`DeltaT = (-1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = -7.2^@"C"`

which results in

`T_"final" = - 7.2^@"C" + 98.8^@"C" = 91.6^@"C" ->` good