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Chapter 6: Problem 58

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 0.137 J/(g°C), and the molar heat capacity is 27.48 J/(mol°C).

Step by step solution

01

Identify the given information

We are given the following information: Q = 585 J m = 125.6 g T1 = 20.0°C T2 = 53.5°C M (molar mass of mercury) = 200.59 g/mol
02

Calculate the temperature change (ΔT)

To find ΔT, subtract the initial temperature (T1) from the final temperature (T2): ΔT = T2 - T1 ΔT = 53.5°C - 20.0°C ΔT = 33.5°C
03

Calculate the specific heat capacity (c)

Use the formula Q = mcΔT to find the specific heat capacity (c): 585 J = (125.6 g)(c)(33.5°C) c = 585 J / (125.6 g × 33.5°C) c = 0.137 J/(g°C)
04

Calculate the moles of mercury

To find the moles (n) of mercury, divide the mass (m) by its molar mass (M): n = m / M n = 125.6 g / 200.59 g/mol n = 0.626 mol
05

Calculate the molar heat capacity (C)

Now that we have the specific heat capacity (c) and the moles of mercury (n), we can calculate the molar heat capacity (C) using the formula: C = c × M C = 0.137 J/(g°C) × 200.59 g/mol C = 27.48 J/(mol°C) So, the specific heat capacity of mercury is 0.137 J/(g°C), and the molar heat capacity is 27.48 J/(mol°C).

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Most popular questions from this chapter

The best solar panels currently available are about 19\(\%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1 \text { kilowatt }\) hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 132 for the energy rate supplied by the sun.)

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