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

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Answer 1 · 2016-11-03T06:01:45

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

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 · 2016-11-03T06:49:56

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

Use the specific heat equation:

![http://pinstopin.com/specific-heat-equation]( useruploads.socratic.org )

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.

$Delta T+T_i=T_f$

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