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Chapter 6: Problem 59

A 30.0 -g sample of water at \(280 . \mathrm{K}\) is mixed with 50.0 g water at \(330 . \mathrm{K}\) . Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Short Answer

Expert verified
The final temperature of the water mixture is \(311.25\:K\).

Step by step solution

01

Write the given information

Mass of water sample 1, \(m_1 = 30.0\:g\) Initial temperature of water sample 1, \(T_1 = 280\:K\) Mass of water sample 2, \(m_2 = 50.0\:g\) Initial temperature of water sample 2, \(T_2 = 330\:K\) The specific heat capacity of water, \(c = 4.184\:J/(g\:K)\)
02

Calculate the heat transfer for both water samples

Let's assume the final temperature of the mixture is \(T_f\). For water sample 1, the heat transfer will be: \(Q_1 = m_1c\Delta T_1 = m_1c(T_f - T_1)\) For water sample 2, the heat transfer will be: \(Q_2 = m_2c\Delta T_2 = m_2c(T_2 - T_f)\) Since heat lost by water sample 2 is equal to the heat gained by water sample 1 and there's no heat loss to surroundings, we can write the equation as: \(Q_1 = Q_2\)
03

Solve for the final temperature

Plug in the values of \(Q_1\) and \(Q_2\) in the equation \(Q_1 = Q_2\): \(m_1c(T_f - T_1) = m_2c(T_2 - T_f)\) Divide both sides by \(c\): \(m_1(T_f - T_1) = m_2(T_2 - T_f)\) Now, solve for \(T_f\): \(T_f(m_1 + m_2) = m_1T_1 + m_2T_2\) \(T_f = \frac{m_1T_1 + m_2T_2}{m_1 + m_2}\) Plug in the values: \(T_f = \frac{(30.0\:g)(280\:K) + (50.0\:g)(330\:K)}{30.0\:g + 50.0\:g}\) \(T_f = \frac{8400 + 16500}{80}\) \(T_f = \frac{24900}{80}\) \(T_f = 311.25\:K\)
04

State the final temperature

The final temperature of the water mixture is \(311.25\:K\).

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Most popular questions from this chapter

Some automobiles and buses have been equipped to burn propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Compare the amounts of energy that can be obtained per gram of \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) and per gram of gasoline, assuming that gasoline is pure octane, $\mathrm{C}_{8} \mathrm{H}_{18}(l) .\( See Example \)6.11 .$ ) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

At 298 \(\mathrm{K}\) , the standard enthalpies of formation for $\mathrm{C}_{2} \mathrm{H}_{2}(g)$ and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are 227 \(\mathrm{kJ} / \mathrm{mol}\) and \(49 \mathrm{kJ} / \mathrm{mol},\) respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. the standard enthalpy of combustion of liquid ethanol, $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) . $$ c. the standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)$ e. the enthalpy of combustion of liquid benzene, $\mathrm{C}_{6} \mathrm{H}_{6}(l)$ f. the enthalpy of solution of solid ammonium bromide.

You have a 1.00 -mole sample of water at \(-30 .^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\) . Calculate \(q\) for the entire process. Use the following data. $$ \begin{aligned} \text { Specific heat capacity of ice } &=2.03 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of water } &=4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of steam } &=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \end{aligned} $$ $$ \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{fision}}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) $$ $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{vaporization}}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 100 .^{\circ} \mathrm{C}\right) $$

The bomb calorimeter in Exercise 112 is filled with 987 \(\mathrm{g}\) water. The initial temperature of the calorimeter contents is $23.32^{\circ} \mathrm{C} .\( A \)1.056-\mathrm{g}\( sample of benzoic acid \)\left(\Delta E_{\mathrm{comb}}=\right.\( \)-26.42 \mathrm{kJ} / \mathrm{g}$ ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?
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