Question

Question #2c289

Answer 1

`"23,000 J"`

Explanation:

Your tool of choice here will be the equation

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

Here

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

In your case, the change in temperature is equal to `66^@"C"`. In other words, the difference between the final and the initial temperatures of the water is equal to `66^@"C"`.

Now, look up the specific heat for water. You should find it listed as

`c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)`

The specific heat of a substance is a measure of how much energy is needed to increase the temperature of `"1 g"` of said substance by `1^@"C"`.

So, plug in your values to find

`q = 83 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 66color(red)(cancel(color(black)(""^@"C")))`

`q = color(darkgreen)(ul(color(black)("23,000 J")))`

The answer must be rounded to two sig figs, the number of significant figures you have for the mass of the sample and for the change in temperature.