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Chapter 6: Problem 58

Hydrogen gives off \(120 . \mathrm{J} / \mathrm{g}\) of energy when burned in oxygen, and methane gives off \(50 . \mathrm{J} / \mathrm{g}\) under the same circumstances. If a mixture of \(5.0 \mathrm{~g}\) hydrogen and \(10 . \mathrm{g}\) methane is burned, and the heat released is transferred to \(50.0 \mathrm{~g}\) water at \(25.0^{\circ} \mathrm{C}\), what final temperature will be reached by the water?

Short Answer

Expert verified
The final temperature of the water can be found by following the steps provided in the solution. First, calculate the total heat released from the hydrogen and methane mixture: Q = Q_h + Q_m = \(5.0g × 120 \frac{J}{g} + 10.0g × 50 \frac{J}{g}\) = 600 J + 500 J = 1100 J Next, calculate the temperature change in the water using the equation: ΔT = \(\frac{Q}{m_{water} × C_{water}}\) = \(\frac{1100J}{50.0g × 4.18 \frac{J}{g °C}}\) = 5.26°C Finally, find the final temperature of the water: T_final = T_initial + ΔT = 25.0°C + 5.26°C = 30.26°C

Step by step solution

01

Calculate the total heat released from the hydrogen and methane mixture

To do this, we'll multiply the mass of each gas by its respective energy release per gram. Total heat released (Q) = Heat released from hydrogen (Q_h) + Heat released from methane (Q_m) First, we calculate the heat released by the hydrogen: Q_h = mass of hydrogen × energy released per gram of hydrogen Q_h = \(5.0g × 120 \frac{J}{g}\) Then, we calculate the heat released by the methane: Q_m = mass of methane × energy released per gram of methane Q_m = \(10.0g × 50 \frac{J}{g}\) Finally, we add the heats released by hydrogen and methane to get the total heat released.
02

Calculate the total heat released from the mixture

Now that we know the heat released by hydrogen and methane, we can find the total heat released by the mixture: Q = Q_h + Q_m = \(5.0g × 120 \frac{J}{g} + 10.0g × 50 \frac{J}{g}\)
03

Calculate the temperature change in the water

To find the temperature change in the water, we'll use the equation: Q = m_water × C_water × ΔT where Q is the heat absorbed by the water, m_water is the mass of the water, C_water is the specific heat capacity of water (4.18 J/g°C), and ΔT is the temperature change in the water. We can rearrange this equation to solve for ΔT: ΔT = \(\frac{Q}{m_{water} × C_{water}}\)
04

Calculate the final temperature of the water

Now, we can find the final temperature of the water (T_final) by adding the temperature change (ΔT) to the initial temperature of the water (T_initial): T_final = T_initial + ΔT = 25.0°C + ΔT Plug in the values for the temperature change that we calculated in step 3.

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Most popular questions from this chapter

Consider the following equations: $$ \begin{array}{rlrl} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & & \Delta H=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \Delta H & =-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \mathrm{A} & & \Delta H= & 64.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2 , and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

A \(5.00-\mathrm{g}\) sample of aluminum pellets (specific heat capacity \(=\) \(0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity \(=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) are heated to \(100.0^{\circ} \mathrm{C}\). The mixture of hot iron and aluminum is then dropped into \(97.3 \mathrm{~g}\) water at \(22.0^{\circ} \mathrm{C}\). Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

In which of the following systems is(are) work done by the surroundings on the system? Assume pressure and temperature are constant. a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) b. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) c. \(4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{array}{cr} \text { Equation } & \Delta H(\mathrm{k}\rfloor) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(I) & -1010 \\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -143 \\ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(I) & -286 \end{array} $$

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?
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