Question

Question #d4a35

Answer 1

`"12.8 kJ"`

Explanation:

The key to this problem lies with the value of water's specific heat, which is said to be equal to

`c_w = "4.184 J g"^(-1)""^@"C"^(-1)`

Now, a substance's specific heat tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In your case, water's specific heat tells you that you need `"4.184 J"` of energy in the form of heat in order to increase the temperature of `"1 g"` of water by `1^@"C"`.

Notice that specific heat is expressed in joules per gram per degree Celsius. This means that you can break up the calculation into two parts

• determine how much heat you need to add in order to increase the temperature of `"45.0 g"` of water by `1^@"C"`

• use this value to determine how much heat would be needed in order to increase the sample's temperature by `DeltaT`

In this case, `DeltaT`, which represents change in temperature, will be equal to

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

`DeltaT = 74.0^@"C" - 6.2^@"C" = 67.8^@"C"`

So, to increase the temperature of `"45.0 g"` of water by `1^@"C"`, you'd need

`45.0 color(red)(cancel(color(black)("g"))) * overbrace(("4.184 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed per 1"^@"C")) = "188.28 J"^@"C"^(-1)`

If this is how much heat you need to produce a `1^@"C"` increase in temperature for a `"45.0-g"` sample of water, it follows that a `67.8^@"C"` increase in temperature would require

`67.8 color(red)(cancel(color(black)(""^@"C"))) * "188.28 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "12,765.4 J"`

I'll leave the answer rounded to three sig figs and express it in kilojoules

`"heat needed" = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))`

ALTERNATIVE APPROACH

You can also solve this problem by using the following equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values to get

`q = 45.0 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 67.8color(red)(cancel(color(black)(""^@"C")))`

`q = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))`