# Question 769b2

Jul 25, 2017

$\text{270 J}$

#### Explanation:

The key here is the value of water's specific heat, which, as you know, tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

${c}_{\text{water" = "4.184 J g"^(-1)""^@"C}}^{- 1}$

You can thus say that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to supply it with $\text{4.184 J}$ of heat.

Now, you're dealing with $\text{1.0432 g}$ of water, so start by calculating the amount of heat needed to increase the temperature of this much water by ${1}^{\circ} \text{C}$.

1.0432 color(red)(cancel(color(black)("g"))) * overbrace("4.184 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C"))^(color(blue)("the specific heat of water")) = "4.365 J"""^@"C"^(-1)

So, you now know that in order to increase the temperature of $\text{1.0432 g}$ of water by ${1}^{\circ} \text{C}$, you need $\text{4.365 J}$ of heat.

But since you want to increase the temperature of the sample by

${88}^{\circ} \text{C" - 25.0^@"C" = 63^@"C}$

you can say that you will need a total of

63 color(red)(cancel(color(black)(""^@"C"))) * overbrace("4.365 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 1.0432 g of water")) = "274.995 J"#

Rounded to two sig figs, the number of sig figs you have for the final temperature of the water, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat needed = 270 J}}}}$