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Chapter 7: Problem 57

A \(150.0 -\mathrm{g}\) sample of a metal at \(75.0^{\circ} \mathrm{C}\) is added to \(150.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(15.0^{\circ} \mathrm{C}\). The temperature of the water rises to \(18.3^{\circ} \mathrm{C}\). Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

Short Answer

Expert verified
The specific heat capacity of the metal can be calculated using the formula \( c_m = \frac{m_w \cdot c_w \cdot ΔT_w}{m_m \cdot (T_{m_i} - T_{m_f})} \) Plugging in the known values: \( c_m = \frac{150.0\,\text{g} \cdot 4.18\,\text{J/g°C} \cdot (18.3\,°\text{C} - 15.0\,°\text{C})}{150.0\,\text{g} \cdot (75.0\,°\text{C} - 18.3\,°\text{C})} \) Solving the equation, we find that the specific heat capacity of the metal is approximately \(0.22\,\text{J/g°C}\).

Step by step solution

01

Recognizing heat transfer concept

Since the amount of heat lost by the metal sample is gained by the water sample, we can express the heat transfer using the following formula: \(Q_{lost} = Q_{gained}\) We also know that the formula for the heat transfer is \(Q = m \cdot c \cdot ΔT\) where \(Q\) represents the heat transfer, \(m\) represents the mass, \(c\) represents the specific heat capacity, and \(ΔT\) represents the temperature difference. Tag_titleStep 2: Identify the known values
02

The information given in the problem is as follows: Mass of the metal sample (\(m_m\)) = 150.0 g Initial temperature of the metal sample (\(T_{m_i}\)) = 75.0°C Mass of the water sample (\(m_w\)) = 150.0 g Initial temperature of the water sample (\(T_{w_i}\)) = 15.0°C Final temperature of the water sample (\(T_{w_f}\)) = 18.3°C

Step 3: Calculate the temperature changes for the metal and water samples
03

We need to find the temperature difference for both the metal and water samples: For the metal: \(ΔT_m = T_{m_i} - T_{m_f}\) For the water: \(ΔT_w = T_{w_f} - T_{w_i}\)

Step 4: Determine the heat gained by the water
04

We'll first find the heat gained by the water using the formula \(Q = m \cdot c \cdot ΔT\), where the specific heat capacity of water (\(c_w\)) is 4.18 J/g°C: \(Q_{gained} = m_w \cdot c_w \cdot ΔT_w\)

Step 5: Calculate the specific heat capacity of the metal
05

Now, using the fact that heat lost is equal to heat gained (\(Q_{lost} = Q_{gained}\)), and the formula \(Q_{lost} = m_m \cdot c_m \cdot ΔT_m\), we can calculate the specific heat capacity of the metal: \( m_m \cdot c_m \cdot (T_{m_i} - T_{m_f}) = m_w \cdot c_w \cdot ΔT_w \)

Solving for \(c_m\), the specific heat capacity of the metal, remembering that \(ΔT_m = T_{m_i} - T_{m_f}\), we get: \( c_m = \frac{m_w \cdot c_w \cdot ΔT_w}{m_m \cdot (T_{m_i} - T_{m_f})} \) Finally, plug in the known values and solve for the specific heat capacity of the metal.

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Most popular questions from this chapter

The reaction $$\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is \(-227 \mathrm{kJ}\). In designing a sulfuric acid plant, is it necessary to provide for heating or cooling of the reaction mixture? Explain.

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. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)

The standard enthalpy of combustion of ethene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\) is \(-1411.1 \mathrm{kJ} / \mathrm{mol}\) at \(298 \mathrm{K}\). Given the following enthalpies of formation, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\). $$\begin{array}{ll}\mathrm{CO}_{2}(g) & -393.5 \mathrm{kJ} / \mathrm{mol} \\\\\mathrm{H}_{2} \mathrm{O}(l) & -285.8 \mathrm{kJ} / \mathrm{mol}\end{array}$$

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{kg}\) water and has a total heat capacity of \(10.84 \mathrm{kJ} /^{\circ} \mathrm{C},\) what is the heat capacity of the calorimeter components?

Hydrogen gives off \(120 .\) J/g of energy when burned in oxygen, and methane gives off \(50 .\) J/g under the same circumstances. If a mixture of 5.0 g hydrogen and \(10 .\) 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?
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