# Question 3269e

Jul 8, 2017

#### Answer:

Here's what I got.

#### Explanation:

I'm going to assume that you're interested in determining the amount of heat needed to convert $\text{89.5 g}$ of liquid water at its normal melting point of ${0}^{\circ} \text{C}$ to liquid water at its normal boiling point of ${100}^{\circ} \text{C}$.

In other words, I will assume that you don't have to go from solid water at ${0}^{\circ} \text{C}$ to water vapor at ${100}^{\circ} \text{C}$, i.e. that no phase change is involved here.

Now, the specific heat of water, which tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, is equal to ${\text{4.18 J g"^(-1)""^@"C}}^{- 1}$. This means that in order to increase the temperature of $\text{1 g}$ of liquid water by ${1}^{\circ} \text{C}$, you need to provide it with $\text{4.18 J}$ of heat.

In your case, you would need

89.5 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "374.11 J"""^@"C"^(-1)

in order to increase the temperature of your sample of water. This value tells you that every ${1}^{\circ} \text{C}$ increase in the temperature of your sample requires $\text{374.11 J}$ of heat.

You can thus say that a change in temperature of

${100}^{\circ} \text{C" - 0^@"C" = 100^@"C}$

will require

100 color(red)(cancel(color(black)(""^@"C"))) * overbrace("374.11 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 89.5 g of water")) = "37,411 J"#

of heat. Rounded to three sig figs, the number of significant figures you have for the mass of water, and expressed in kilojoules, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat needed = 37.4 kJ}}}}$