/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Hydrogen gives off \(120 . \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(30.24 ^{\circ}\mathrm{C}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is a fascinating concept that describes the movement of energy between substances due to a difference in temperature. When we burn substances like hydrogen and methane, they release energy in the form of heat. This heat then transfers to another substance---in our case, water.

In calorimetry problems, we often deal with heat transfer from a chemical reaction to a surrounding substance, often resulting in a change in temperature. This process consists of:
  • Exothermic reactions that give off heat (like our burning fuel),
  • Endothermic reactions that absorb heat.
Heat transfer is directional; it always moves from the hotter object to a cooler one. This fundamental principle helps us predict resultant changes in temperature and energy distribution between substances.
Specific Heat Capacity
Specific heat capacity is a property that describes how much energy is required to raise the temperature of one gram of a substance by one degree Celsius. Different materials require varying amounts of energy to change their temperature.

For water, this amount is notably high, about 4.18 J/g°C. This means that water can absorb or release a lot of energy with only a small change in temperature. In our exercise, understanding water's specific heat capacity allows us to determine how much its temperature will change when it absorbs the heat released from burning fuels.

Calculating temperature changes with specific heat capacity involves the formula:
  • \[ Q = mc\Delta T \]
  • Here: \( Q \) is heat energy, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change.
This ability to calculate changes is key to predicting outcomes in chemical reactions and heat transfer scenarios.
Enthalpy of Combustion
The enthalpy of combustion is the energy released as heat when a substance combusts in adequate oxygen conditions. It's usually expressed in Joules per gram (J/g) or kilojoules per mole (kJ/mol).

In our given problem, the enthalpy of combustion is provided for both hydrogen and methane:
  • Hydrogen releases 120 J/g,
  • Methane releases 50 J/g.
Understanding this property allows us to calculate how much energy the burning of these fuels releases. By knowing the mass of each fuel, we compute the total energy that affects other substances like water. This data is crucial to solving calorimetry and heat transfer problems effectively.
Temperature Change
Temperature change is a critical outcome of heat transfer. It's what we ultimately seek to understand in calorimetry problems: how does the temperature of a substance like water change as it absorbs or expels heat?

Using the formula \( \Delta T = \frac{Q}{mc} \), we translate the total energy absorbed (1100 J in our problem) and specific heat capacity to find out how much the temperature of water changes. Calculations showed:
  • Initial water temperature: 25.0°C,
  • Energy absorbed: 1100 J,
  • Resulting Temperature Change: ≈ 5.24 °C.
Therefore, the final temperature was calculated to be around 30.24°C. This kind of precise change depends on a precise understanding of thermodynamic principles and careful computational steps.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It has been determined that the body can generate 5500 \(\mathrm{kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is 40.6 \(\mathrm{kJ} / \mathrm{mol.} )\)

In a coffee-cup calorimeter, 1.60 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with 75.0 \(\mathrm{g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\) . After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\) . Assuming the solution has a heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of kJ/mol.

A piston performs work of \(210 . \mathrm{L} \cdot\) atm on the surroundings, while the cylinder in which it is placed expands from \(10 . \mathrm{L}\) to 25 \(\mathrm{L}\) . At the same time, 45 \(\mathrm{J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?

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.

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for \(\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\) \((-2755 \mathrm{kJ} / \mathrm{mol}),\) and \(\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\) calculate \(\Delta H\) for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.