Question

4790 joules of heat are added to 222 grams of water originally at 27.7 C. What is the final temperature of the water?

Answer 1

If I am correct, you will need to know the equation `Q=mc ΔT` to solve this problem.

Explanation:

Here, you are given the values for Q, m and c can be obtained from your textbook. According to mine, the value should be 4.19 J/C.

`4790 J= 222g * 4.19 * (x°C-27.7°C)`

Note that I let x represent the final temperature. And now, all you have to do is some simple arithmetic and solve for x!

The final answer is 32.8°C

Answer 2

The final temperature is `"32.9"^@"C"`.

Explanation:

Use the specific heat equation:

Known

`Q="4790 J"`
`c_"water"=4.184 "J"/("g·"^@"C")`
`m="222 g"`

`T_"initial"="27.7"^@"C"`

Unknown

`T_"final"`

Solution

Rearrange the equation to isolate `Delta T`.

`Q=cmDeltaT`

Divide `Q` by `cm`.

`Delta T=Q/(cm)`

`Delta T=(4790cancel"J")/((4.184cancel"J")/(cancel"g·"^@"C")xx222cancel"g"` `=` `"5.16"^@"C"`

Now that `Delta T` is known, we'll write it as equal to the final temperature minus the initial temperature.

`Delta T=T_f-T_i`

Add `T_1` to both sides.

`Delta T+T_i=T_f`

`5.16^@"C"+"27.7"^@"C"="32.9"^@"C"`