# Question 087c4

May 14, 2017

${91.6}^{\circ} \text{C}$

#### Explanation:

For starters, you need to make a note that the specific heat of water is equal to

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

This means that when the temperature of $\text{1 g}$ of liquid water decreases by ${1}^{\circ} \text{C}$, $\text{1 cal}$ of heat is being given off.

Now, your tool of choice here will be this equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat gained by the water
• $m$ is the mass of the water
• $c$ is the specific heat of water
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You need to solve for $\Delta T$, so rearrange the equation as

$\Delta T = \frac{q}{m \cdot c}$

Plug in your values to find

DeltaT = (1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = 7.2^@"C"

Now, you know that the sample is losing heat, so you can say that the final temperature will be ${7.2}^{\circ} \text{C}$ lower than the initial temperature

${T}_{\text{final" = T_"initial}} - \Delta T$

which, in your case, will be equal to

T_"final" = 98.8^@"C" - 7.2^@"C" = color(darkgreen)(ul(color(black)(91.6^@"C")))

The answer is rounded to one decimal place.

$\textcolor{w h i t e}{a}$

SIDE NOTE It's important to notice that if you use

$\Delta T = {T}_{\text{final" - T_"initial}}$

you get

${T}_{\text{final" = DeltaT + T_"initial}}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{T}_{\text{final" = 7.2^@"C" + 98.8^@"C" = 106^@"C}}}}} \to$ not good

The problem here is that the sample is actually giving off heat, so you should use

$q = - \text{1794 cal}$

in the equation. The minus sign is needed because the final temperature is lower than the initial temperature. This will get you

DeltaT = (-1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = -7.2^@"C"#

which results in

${T}_{\text{final" = - 7.2^@"C" + 98.8^@"C" = 91.6^@"C}} \to$ good