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Chapter 6: Problem 52

It takes \(585 \mathrm{~J}\) of energy to raise the temperature of \(125.6 \mathrm{~g}\) mercury from \(20.0^{\circ} \mathrm{C}\) to \(53.5^{\circ} \mathrm{C}\). Calculate the specific heat capacity and the molar heat capacity of mercury.

Short Answer

Expert verified
The specific heat capacity of mercury is \(0.139 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{掳C}}\), and the molar heat capacity is approximately \(27.88 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{掳C}}\).

Step by step solution

01

Identify the given information and the formula to use

We are given the following information: - Energy (Q) = 585 J - Mass (m) = 125.6 g - Initial temperature (Ti) = 20.0掳C - Final temperature (Tf) = 53.5掳C Our goal is to find the specific heat capacity (c) and the molar heat capacity. We will use the formula for heat transfer: Q = mc螖T, where Q is the energy/heat, m is the mass, c is the specific heat capacity, and 螖T is the change in temperature.
02

Calculate the change in temperature (螖T)

螖T = Tf - Ti 螖T = 53.5掳C - 20.0掳C 螖T = 33.5掳C
03

Use the heat transfer formula to find the specific heat capacity (c)

Rearrange the formula to solve for c: c = Q / (m * 螖T) Plug in the given values: c = 585 J / (125.6 g * 33.5掳C) c = 585 J / 4206.2 g掳C c = 0.139 \(\frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{掳C}}\) So, the specific heat capacity of mercury is 0.139 J/g掳C.
04

Calculate the molar heat capacity

To find the molar heat capacity, we need the molar mass of mercury (Hg). The molar mass of Hg is 200.59 g/mol. Molar Heat Capacity = Specific Heat Capacity * Molar Mass Molar Heat Capacity = 0.139 \(\frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{掳C}}\) * 200.59 \(\frac{\mathrm{g}}{\mathrm{mol}}\) Molar Heat Capacity 鈮� 27.88 \(\frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{掳C}}\) The molar heat capacity of mercury is approximately 27.88 J/mol掳C.

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Most popular questions from this chapter

A system undergoes a process consisting of the following two steps: Step 1: The system absorbs \(72 \mathrm{~J}\) of heat while \(35 \mathrm{~J}\) of work is done on it. Step 2: The system absorbs \(35 \mathrm{~J}\) of heat while performing \(72 \mathrm{~J}\) of work. Calculate \(\Delta E\) for the overall process.

Nitromethane, \(\mathrm{CH}_{3} \mathrm{NO}_{2}\), can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ a. The standard enthalpy change of reaction \(\left(\Delta H_{\mathrm{rxn}}^{\circ}\right)\) for the balanced reaction (with lowest whole- number coefficients) is \(-1288.5 \mathrm{~kJ} .\) Calculate the \(\Delta H_{\mathrm{f}}^{\circ}\) for nitromethane. b. A \(15.0\) - \(\mathrm{L}\) flask containing a sample of nitromethane is filled with \(\mathrm{O}_{2}\) and the flask is heated to \(100 .^{\circ} \mathrm{C}\). At this temperature, and after the reaction is complete, the total pressure of all the gases inside the flask is 950 . torr. If the mole fraction of nitrogen ( \(\chi_{\text {nitrogen }}\) ) is \(0.134\) after the reaction is complete, what mass of nitrogen was produced?

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(1.00 \mathrm{~atm}\) and \(100 .^{\circ} \mathrm{C}\) occupies a volume of \(30.6 \mathrm{~L}\). When one mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to one mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{~g} / \mathrm{cm}^{3}\), calculate \(\Delta E\) for the condensation of one mole of water at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}\).

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5\) L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{aligned} 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1010 . \mathrm{kJ} \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-317 \mathrm{~kJ} \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-143 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \Delta H &=-286 \mathrm{~kJ} \end{aligned} $$
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