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Chapter 7: Problem 50

The specific heat capacity of silver is \(0.24 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{g}\) Ag from \(273 \mathrm{K}\) to \(298 \mathrm{K}\). b. Calculate the energy required to raise the temperature of 1.0 mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample silver.

Short Answer

Expert verified
a. The energy required to raise the temperature of \(150.0 \, g\) Ag from \(273 \, K\) to \(298 \, K\) is \(900 \, J\). b. The molar heat capacity of silver is \(25.89 \, J/mol \cdot ^{o}C\). c. The mass of the sample silver is \(21708.3 \, g\).

Step by step solution

01

Part a: Calculate the energy required for a 150 g silver sample

The given data are: - mass (m): \(150.0 \, g\) - specific heat capacity (c): \(0.24 \, J/g \cdot ^{o}C\) - temperature change (\(\Delta T\)): \(298 \, K - 273 \, K = 25 \, ^{o}C\) Now, we will use the formula \(q = mc\Delta T\). \(q = (150.0 \, g) \times (0.24 \, J/g \cdot ^{o}C) \times (25 \, ^{o}C) = 900 \, J\) Therefore, it takes 900 Joules of energy to raise the temperature of \(150.0 \, g\) Ag from \(273 \, K\) to \(298 \, K\).
02

Part b: Calculate molar heat capacity of silver

First, we need to find the molar mass of Silver (Ag): - Given, 1 mole Ag has a molar mass of \(107.87 \, g\). Now, we need to find the energy required to raise the temperature of 1 mole of Ag by \(1.0 \, ^{o}C\). We use the same formula, \(q = mc\Delta T\), but with the mass of 1 mole Ag. m = \(107.87 \, g\) \(\Delta T = 1 \, ^{o}C\) \(q_{1 mole} = (107.87 \, g) \times (0.24 \, J/g \cdot ^{o}C) \times (1 \, ^{o}C) = 25.89 \, J\) Therefore, the molar heat capacity of silver is \(25.89 \, J/mol \cdot ^{o}C\).
03

Part c: Calculate the mass of the silver sample

The given data are: - energy (q): \(1.25 \, kJ = 1250 \, J\) - temperature change (\(\Delta T\)): \(15.2 \, ^{o}C - 12.0 \, ^{o}C = 3.2 \, ^{o}C\) - specific heat capacity (c): \(0.24 \, J/g \cdot ^{o}C\) We will again use the formula \(q = mc\Delta T\), but this time, we will need to solve for mass (m). \(1250 \, J = m \times (0.24 \, J/g \cdot ^{o}C) \times (3.2 \, ^{o}C)\) Now, we will isolate the mass (m) variable by dividing both sides of the equation by \(0.24 \, J/g \cdot ^{o}C\) and \(3.2\, ^{o}C\). \(m = \frac{1250 \, J}{(0.24 \, J/g \cdot ^{o}C) \times (3.2 \, ^{o}C)} = 21708.3 \, g\) Therefore, the mass of the sample silver is \(21708.3 \, g\).

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Most popular questions from this chapter

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

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