# Question 11e6e

Feb 22, 2017

${43}^{\circ} \text{C}$

#### Explanation:

The idea here is that you can assume that the heat lost by the iron bar as it cools will be equal to the heat gained by the water as it warms.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{- {q}_{\text{iron" = q_"water}}}}}$

The minus sign is used here because heat lost is, by definition, negative.

Your tool of choice here will be the 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 lost or gained by the substance
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

You should know that the specific heat of water is

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

If you take ${T}_{f}$ to be the final temperature reached by the iron bar + water system, you can say that you have

• For the iron bar

q_"iron" = 45.0 color(red)(cancel(color(black)("g"))) * "0.11 cal" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 100)color(red)(cancel(color(black)(""^@"C")))

${q}_{\text{iron" = 4.95 * (T_f - 100)" cal}}$

• For the water

q_"water" = 90.0 color(red)(cancel(color(black)("g"))) * "1 cal" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 40)color(red)(cancel(color(black)(""^@"C")))#

${q}_{\text{water" = 90.0 * (T_f - 40)" cal}}$

Since the heat lost by the iron bar must be equal to heat gained by the water, you will have -- do not forget about the minus sign used for the heat lost by the iron bar!

$- 4.95 \cdot \left({T}_{f} - 100\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{cal"))) = 90.0 * (T_f - 40)color(red)(cancel(color(black)("cal}}}}$

This will get you

$- 4.95 \cdot {T}_{f} + 495 = 90.0 \cdot {T}_{f} - 3600$

$94.95 \cdot {T}_{f} = 4095 \implies {T}_{f} = \frac{4095}{94.95} = 43.13$

Therefore, you can say that the iron bar will cool to a final temperature of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{T}_{f} = {43}^{\circ} \text{C}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the starting temperatures of the iron bar and of the water.