# Question #d4a35

##### 1 Answer

#### Explanation:

The key to this problem lies with the value of water's **specific heat**, which is said to be equal to

#c_w = "4.184 J g"^(-1)""^@"C"^(-1)#

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

In your case, water's specific heat tells you that you need

Notice that specific heat is expressed in *joules per gram per degree Celsius*. This means that you can break up the calculation into two parts

determine how much heat you need to add in order to increase the temperature of#"45.0 g"# of water by#1^@"C"#

use this value to determine how much heat would be needed in order to increase the sample's temperature by#DeltaT#

In this case,

#DeltaT = T_"final" - T_"initial"#

#DeltaT = 74.0^@"C" - 6.2^@"C" = 67.8^@"C"#

So, to increase the temperature of

#45.0 color(red)(cancel(color(black)("g"))) * overbrace(("4.184 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed per 1"^@"C")) = "188.28 J"^@"C"^(-1)#

If this is how much heat you need to produce a

#67.8 color(red)(cancel(color(black)(""^@"C"))) * "188.28 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "12,765.4 J"#

I'll leave the answer rounded to three **sig figs** and express it in *kilojoules*

#"heat needed" = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#

**ALTERNATIVE APPROACH**

You can also solve this problem by using the following 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**

Plug in your values to get

#q = 45.0 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 67.8color(red)(cancel(color(black)(""^@"C")))#

#q = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#