# The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

Oct 26, 2015

$\text{239 J}$

#### Explanation:

First thing first, you mistyped the specific heat of water, which should be

c_"water" = 4.18"J"/("g" ""^@"C")

Now, a substance's specific heat tells you how much heat is required to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In the case of water, you would need $\text{4.18 J}$ to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

Notice that your sample of water has a mass of $\text{1 g}$ as well, which means that the only factor that will determine the amount of heat needed will be the difference in temperature.

The equation that establishes a relationshop between heat and change in temperature looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed
$c$ - the specific heat of the substance, in your case of water
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

Plug in your values and solve for $q$ to get

$q = 1.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (83.7 - 26.5)color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{239.096 J}$

Rounded to three sig figs, the answer will be

$q = \textcolor{g r e e n}{\text{239 J}}$

SIDE NOTE Notice how much heat would be needed to increase the temperature of 57.2 g of water by ${1}^{\circ} \text{C}$. $\Delta T$ will now be equal to ${1}^{\circ} \text{C}$ and you'd have

$q = 57.2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 1color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{239 J}$

This tells you that you need the same amount of heat to increase the temperature of $\text{1 g}$ of water by ${57.2}^{\circ} \text{C}$, as you do to increase the temperature of $\text{57.2 g}$ of water by ${1}^{\circ} \text{C}$.