What is the temperature change in 224 g of water upon the absorption of 55 kJ of heat, the specific heat of water is 4.18 J/g °C?

1 Answer
Dec 18, 2016

DeltaT = 58.7^oC

Explanation:

To obtain the change in temperature, let's use the equation below:
http://slideplayer.com/slide/4758097/http://slideplayer.com/slide/4758097/

In this question, heat has units of kJ, which does not match up with the units given in the image above. So what we can do first is convert kJ to J by multiplying that value of q by 1000 using the conversion factor below:
http://converters360.com/energy/kilojoule-to-joule-conversion.htmhttp://converters360.com/energy/kilojoule-to-joule-conversion.htm

Here are the variables that we know:

Q = 55cancel"kJ"xx(1000J)/(1cancel"kJ") = 55,000 J
m = 224 g
C = 4.18J/(gxx^oC)

All we have to do is rearrange the equation to solve for Delta T. This can be accomplished by dividing both sides by C and m to get Delta T by itself like this:

DeltaT = Q/(mxxC)

Now, we just plug in the known values:

DeltaT= (55,000cancelJ)/(224cancelgxx4.18cancelJ/cancel"g"xx^oC)

DeltaT = 58.7^oC