# Question 8b482

Mar 6, 2017

${24}^{\circ} \text{C}$

#### Explanation:

The idea here is that the heat given off by the piece of iron as it cools will be equal to the heat absorbed by the water as it warms.

color(blue)(ul(color(black)(-q_"iron" = q_"water")))" " " "color(darkorange)("(*)")

Keep in mind that we use a minus sign here because, by convention, heat given off carries a negative sign

Before moving forward, look up the specific heat values of water and iron, respectively. You'll find them listed as

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

${c}_{\text{iron" = "0.444 J g"^(-1)""^@"C}}^{- 1}$

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

Now, 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

If you take ${T}_{f}$ to be the final temperature of the iron + water system, you can say that the change in temperature for the piece of iron will be

$\Delta {T}_{\text{iron" = T_f - 95^@"C}}$

Similarly, the change in temperature for the water will be

$\Delta {T}_{\text{water" = T_f - 22^@"C}}$

Convert the two masses from kilograms to grams

$\text{2.4 kg" = 2.4 * 10^(3) color(white)(.)"g" = "2400 g}$

$\text{0.6 kg" = 0.6 * 10^3 color(white)(.)"g" = "600 g}$

You can thus say that the heat given off by the piece of iron will be equal to

q_"iron" = 600 color(red)(cancel(color(black)("g"))) * "0.444 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 95)color(red)(cancel(color(black)(""^@"C")))

${q}_{\text{iron}} = 266.4 \cdot \left({T}_{f} - 95\right)$

The heat absorbed by the water will be equal to

q_"water" = 2400 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 22)color(red)(cancel(color(black)(""^@"C")))#

${q}_{\text{water}} = 10041.6 \cdot \left({T}_{f} - 22\right)$

You can now use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to say that

$- 266.4 \cdot \left({T}_{f} - 95\right) = 10041.6 \cdot \left({T}_{f} - 22\right)$

This will be equivalent to

$- 266.4 {T}_{f} + 25308 = 10041.6 {T}_{f} - 220915.2$

Rearrange to get

$\left(10041.6 {T}_{f} + 225.4\right) \cdot {T}_{f} = 220915.2 + 25308$

You will thus have

${T}_{f} = \frac{220915.2 + 25308}{10041.6 + 226.4} = 23.98$

You can thus say that the final temperature of the iron + water system will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{T}_{f} = {24}^{\circ} \text{C}}}}$

The answer is rounded to two sig figs.