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Chapter 7: Problem 5

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?

Short Answer

Expert verified
The original temperature of the iron was approximately 544.7 degrees Celsius.

Step by step solution

01

Compute the Heat Gained by the Water

Use the formula \( Q = m \cdot C \cdot \Delta T \) to calculate the heat gained by the water, where \( Q \) is the heat, \( m \) is the mass, \( C \) is the specific heat, and \( \Delta T \) is the temperature change. For water, \( m = 375g \), \( C = 4.18 Jg^{-1} ^{\circ} C^{-1} \) (the specific heat of water), and \( \Delta T = 87^{\circ}C - 26^{\circ}C = 61^{\circ}C \). Substituting these values we get \( Q = 375g \cdot 4.18 Jg^{-1} ^{\circ} C^{-1} \cdot 61^{\circ}C = 95722.5 J \).
02

Compute the Temperature Change of the Iron

The heat lost by the iron is equal to the heat gained by the water, so \( Q_{iron} = Q_{water} = 95722.5 J \). We can rearrange the formula from step 1 to find the temperature change of the iron as \( \Delta T = Q / (m \cdot C) \). The mass of the iron is given as 465g and the specific heat is 0.449 Jg^{-1} ^{\circ} C^{-1}, so the temperature change is \( \Delta T = 95722.5 J / (465g \cdot 0.449 Jg^{-1} ^{\circ} C^{-1}) = -457.7^{\circ}C \). The negative sign indicates that the temperature of the iron decreased.
03

Compute the Original Temperature of the Iron

To find the original temperature of the iron, add the magnitude of the temperature change to the final temperature. The final temperature of the iron is the same as the final temperature of the water, which is 87^{\circ}C. So the original temperature of the iron is \( 87^{\circ}C + 457.7^{\circ}C = 544.7^{\circ}C \).

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Most popular questions from this chapter

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

In each of the following processes, is any work done when the reaction is carried out at constant pressure in a vessel open to the atmosphere? If so, is work done by the reacting system or on it? (a) Neutralization of \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) by \(\mathrm{HCl}(\mathrm{aq}) ;\) (b) conversion of gaseous nitrogen dioxide to gaseous dinitrogen tetroxide; (c) decomposition of calcium carbonate to calcium oxide and carbon dioxide gas.

Brass has a density of \(8.40 \mathrm{g} / \mathrm{cm}^{3}\) and a specific heat of \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} . \mathrm{A} 15.2 \mathrm{cm}^{3}\) piece of brass at an initial temperature of \(163^{\circ} \mathrm{C}\) is dropped into an insulated container with \(150.0 \mathrm{g}\) water initially at \(22.4^{\circ} \mathrm{C}\) What will be the final temperature of the brass-water mixture?

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) torches are used in welding. How much heat (in kJ) evolves when 5.0 L of \(C_{2} \mathrm{H}_{2}\) \(\left(d=1.0967 \mathrm{kg} / \mathrm{m}^{3}\right)\) is mixed with a stoichiometric amount of oxygen gas? The combustion reaction is $$\begin{array}{r} \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-1299.5 \mathrm{kJ} \end{array}$$
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