# Question #3d723

##### 1 Answer

#### Explanation:

The equation that establishes a relationship between the amount of heat absorbed by a substance and the resulting *increase in temperature* looks like this

#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**

Notice that you need to know the **mass** of the sample of water, but that that problem provides you with the *volume* of water.

When no mention about water's density is made, you can *assume* it to be equal to

#1.00 color(red)(cancel(color(black)("L"))) * "1 kg"/(1color(red)(cancel(color(black)("L")))) = "1.00 kg"#

Convert this to *grams* by using the conversion factor

#"1 kg" = 10^3"g"#

You will have

#1.00color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.00 * 10^3"g"#

You can find the value of water's **specific heat** here

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

https://en.wikipedia.org/wiki/Heat_capacity#Mass_heat_capacity_of_building_materials

The temperature of the sample increases from

#DeltaT = 85^@"C" - 18^@"C" = 67^@"C"#

Plug in your values to get

#q = 1.00 * 10^3color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 67color(red)(cancel(color(black)(""^@"C")))#

#q = "280,600 J"#

Expressed in *kilojoules* and rounded to two **sig figs**, the answer will be

#q = color(green)(|bar(ul(color(white)(a/a)"281 kJ"color(white)(a/a)|)))#