# Question #71e95

##### 1 Answer

#### Explanation:

In order to be able to solve this problem, you need to know the value of water's **specific heat**, which you'll find listed as

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"water" = "4.18 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#

A substance's *specific heat* tells you how much heat is needed in order to increase the temperature of

In your case, you can say that in order to increase the temperature of

Keep in mind that this much heat must be applied **per gram**, **per degree Celsius**.

Now, you know that your sample of water has a mass of *kilojoules* to *joules* by using the conversion factor

#"1 kJ" = 10^3"J"#

to get

#21.8color(red)(cancel(color(black)("kJ"))) * (10^3"J")/(1color(red)(cancel(color(black)("kJ")))) = "21,800 J"#

At this point, you can determine how much heat would be needed in order to increase the temperature of this sample of water by **per gram** in order to get a

#150. color(red)(cancel(color(black)("g"))) * overbrace("4.18 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed for 1"^@"C increase for 1 g")) = "627 J"#

If you need **difference** between this much heat and the

You can thus say that

#"21,800"color(red)(cancel(color(black)("J"))) * overbrace((1^@"C")/(627color(red)(cancel(color(black)("J")))))^(color(blue)("heat needed for 1"^@"C for 150. g")) = 34.77^@"C"#

If the sample started at an initial temperature of

#T_"final" = 23.4^@"C" + 34.77^@"C" = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

**ALTERNATIVE APPROACH**

You can get the same result by suing the equation

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

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

In your case, you'd have

#DeltaT = q/(m * c)#

#DeltaT = ("21,800" color(red)(cancel(color(black)("J"))))/(150. color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1)) = 34.77^@"C"#

Once again, you'd have

#T_"final" = 23.4^@"C" + DeltaT = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))#