# How much energy will be required to heat a "1.0 kg" mass of water from 25^@"C" to 99^@"C"?

Feb 28, 2015

You'd need $\text{310 kJ}$ to raise the temperature of 1 kg of water from $25$ to ${99}^{\circ} \text{C}$.

So, you have all the information you need to use the equation

$q = m \cdot c \cdot \Delta T$, where

$q$ - the amount of heat needed;
$m$ - the mass of water - in your case 1.0 kg:
$c$ - the specific heat of water;
$\Delta T$ - the difference between the final temperature, ${99}^{\circ} \text{C}$, and the initial temperature, ${25}^{\circ} \text{C}$, of the water.

Plug your data into the equation to get

$q = \text{1,000 g" * 4.18"J"/("g" * ^@"C") * (99-25)^@"C}$

$q = \text{309,320 J" = "+309.3 kJ}$

Rounded to two sig figs, the answer will be

$q = \text{+310 kJ}$

Feb 28, 2015

It will require $\text{310,000 J}$.

#### Explanation:

In order to answer this question, you you will need to use the following equation:

$q = c m \Delta T$,

where $q$ is the quantity of heat gained or lost, $c$ is the specific heat capacity (of water in this case), $m$ is mass in grams, and $\Delta T$ is the difference in temperature, $\Delta T = {T}_{\text{final"-T_"initial}}$

Known/Given:
${c}_{\text{water"= 4.184 "J"/("g"*""^("o")"C}}$

m=1.0color(red)cancel(color(black)("kg"))xx"1000 g"/(1color(red)cancel(color(black)("kg")))=1.0xx10^3 $\text{g}$

${T}_{i} = \text{25"^@"C}$

${T}_{f} = \text{99"^@"C}$

$\Delta T = \text{99"^@"C" - "25"^@"C"="74"^@"C}$

Unknown:
$q$

Solution:

Plug the known values into the equation and solve.

$q = 4.184 \text{J"/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx1.0xx10^3color(red)cancel(color(black)("g"))xx74^@color(red)cancel(color(black)("C")) = "310,000 J}$ (rounded to two significant figures)