# Question 3d723

Mar 25, 2016

#### Answer:

$\text{281 kJ}$

#### 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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the 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 ${\text{1 kg L}}^{- 1}$. This means that you have

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

$\text{1 kg" = 10^3"g}$

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

${c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}$

https://en.wikipedia.org/wiki/Heat_capacityMass_heat_capacity_of_building_materials

The temperature of the sample increases from ${18}^{\circ} \text{C}$ to ${85}^{\circ} \text{C}$, which means that the change in temperature is equal to

$\Delta T = {85}^{\circ} \text{C" - 18^@"C" = 67^@"C}$

Plug in your values to get

$q = 1.00 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 67color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{280,600 J}$

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

$q = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{281 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$