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

2 Answers
Nov 3, 2016

Answer:

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

Nov 3, 2016

Answer:

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

Explanation:

Use the specific heat equation:

http://pinstopin.com/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"#