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Chapter 6: Problem 62

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 will be approximately \(30.24 ^{\circ}\mathrm{C}\).

Step by step solution

01

Calculate the energy released by burning hydrogen

First, we need to find out how much energy is released when 5.0 g of hydrogen is burned. We are given that hydrogen releases 120 J/g when burned in oxygen. Therefore, we can multiply the mass of hydrogen by the energy released per gram to find the total energy released by the hydrogen. Total energy released by hydrogen = (mass of hydrogen) × (energy released per gram) = 5.0 g × 120 J/g = 600 J
02

Calculate the energy released by burning methane

Next, we need to find out how much energy is released when 10 g of methane is burned. We are given that methane releases 50 J/g when burned in oxygen. Therefore, we can multiply the mass of methane by the energy released per gram to find the total energy released by the methane. Total energy released by methane = (mass of methane) × (energy released per gram) = 10 g × 50 J/g = 500 J
03

Calculate the total energy released

Add the energy released by burning hydrogen and methane to find the total energy released. Total energy released = energy from hydrogen + energy from methane = 600 J + 500 J = 1100 J
04

Calculate the temperature change of the water

Now, we'll use the energy released to find the temperature change in the 50.0 g of water. We will use the formula Q = mcΔT, where Q is the heat absorbed (in joules), m is the mass of the substance absorbing the heat (in grams), c is the specific heat capacity of the substance (in J/g°C), and ΔT is the temperature change (in °C). For water, the specific heat capacity is approximately 4.18 J/g°C. We are given the mass of water (50.0 g) and the energy we calculated in Step 3 (1100 J). We can rearrange the formula to solve for ΔT: ΔT = Q / (mc) Temperature change in water = total energy released / (mass of water × specific heat of water) = 1100 J / (50.0 g × 4.18 J/g°C) ≈ 5.24 °C
05

Calculate the final temperature of the water

Now that we have the temperature change in the water, we can add this value to the initial temperature of the water (25.0 °C) to find the final temperature: Final temperature of water = initial temperature + temperature change = 25.0 °C + 5.24 °C ≈ 30.24 °C So the final temperature of the water will be approximately \(30.24 ^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A 0.1964 -g sample of quinone $\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)$ is burned in a bomb calorimeter with a heat capacity of 1.56 \(\mathrm{kJ} / \mathrm{C}\) . The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C}\) . Calculate the energy of combustion of quinone per gram and per mole.

Given: $$ \begin{array}{ll}{2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s)} & {\Delta H^{\circ}=-288 \mathrm{kJ}} \\\ {\mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuO}(s)+\mathrm{Cu}(s)} & {\Delta H^{\circ}=11 \mathrm{kJ}}\end{array} $$ Calculate the standard enthalpy of formation $\left(\Delta H_{f}^{\circ}\right)\( for \)\mathrm{CuO}(s) .$

In a coffee-cup calorimeter, 150.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) HCl is added to 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaOH}\) to make 200.0 \(\mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\) . If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) , calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assume no heat loss to the surroundings.

For the process $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\( at 298 \)\mathrm{K}\( and \)1.0 \mathrm{atm},\( \)\Delta H$ is more positive than \(\Delta E\) by 2.5 \(\mathrm{kJ} / \mathrm{mol}\) . What does the 2.5 \(\mathrm{kJ} / \mathrm{mol}\) quantity represent?

The enthalpy change for the reaction $$ \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?
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