Question 8f7d4

Nov 5, 2017

${47}^{\circ} \text{C}$

Explanation:

If you take ${T}_{f}$ $\text{^@"C}$ to be the final temperature of the sample, you can say that you have

$135 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * (T_f - 21)color(red)(cancel(color(black)(""^@"C"))) = 82.5 color(red)(cancel(color(black)("g"))) * (89 - T_f)color(red)(cancel(color(black)(""^@"C}}}}$

This will get you

${T}_{f} = \frac{82.5 \cdot 89 + 135 \cdot 21}{135 + 82.5}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{T}_{f} = 47}}} \text{ } \to$ rounded to two sig figs

You can thus say that the final temperature of the sample will be ${47}^{\circ} \text{C}$.

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

THE DETAILED VERSION

Assuming that no heat is lost to the surroundings, you can say that the heat lost by the warmer sample will be equal to the heat gained by the colder sample.

color(blue)(ul(color(black)(q_"gained" = - q_"lost")))" "color(darkorange)("(*)")#

The minus sign is used here because heat lost carries a negative sign.

Now, the amount of heat gained/lost depends on the mass of the substance, $m$, on its specific heat, $c$, and on the change in temperature, $\Delta T$, which is calculated by subtracting the initial temperature from the final temperature.

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

If you take ${T}_{f}$ $\text{^@"C}$ to be the final temperature of the sample, you can say that the heat gained by the colder sample will be

${q}_{\text{gained" = "135 g" * c_"water" * (T_f - 21)^@"C}}$

Similarly, the heat lost by the warmer sample will be

${q}_{\text{lost" = "82.5 g" * c_"water" * (T_f - 89)^@"C}}$

Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to say that--do not forget about the minus sign!

$135 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 21) color(red)(cancel(color(black)(""^@"C"))) = - 82.5 color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 89)color(red)(cancel(color(black)(""^@"C}}}}$

This simplifies to

$135 \cdot \left({T}_{f} - 21\right) = - 82.5 \cdot \left({T}_{f} - 89\right)$

which is equivalent to

$135 \cdot \left({T}_{f} - 21\right) = 82.5 \cdot \left(89 - {T}_{f}\right)$

Once again, you will find that the final temperature of the sample will be ${47}^{\circ} \text{C}$.

The answer must be rounded to two sig figs, the number of sig figs you have for the two temperatures.