Thermal Chemistry Calorimeter Problem Answers

1. How much heat will it take to raise the temperature of a 15.0 g gold bracelet from 16.1 °C to 49.3 °C? Assume a constant heat capacity of 25.4 J °C-1mol-1 over this temperature range.

The first thing to look at in this problem is the heat capacity, since its units allow us to set up the problem using dimensional analysis. Next let's look at what we know:

mass of gold = 15.0g o However, we need moles since we are given a molar heat capacity. Moles

of gold = (15.0 g)/(197 g mol-1) = 7.62 x 10-2 mol.

temperature change = 49.3 - 16.1 °C = 33.2°C

Setting up our equation so that the units in the heat capacity cancel to give J = heat:

Heat = (25.4 J °C-1mol-1)(7.62 x 10-2 mol)(33.2 °C) = 64.2 J

Of course we could also plug our values into the heat capacity equation:

Heat = cn T

where c = molar heat capacity, n = moles and T = temperature change in °C:

Heat = (25.4 J °C-1mol-1)(7.62 x 10-2 mol)(33.2 °C) = 64.2 J

2. A 0.95 kg cast iron pan is heated from room temperature (20.0 °C) to 125.0 °C on a gas stove. Assuming the heat capacity of the pan remains constant over this temperature range, calculate the quantity of heat absorbed by the pan (heat capacity of Fe = 25.1 J °C-1mol-1).

For this problem let's start with the heat equation and what we know:

Heat = cn T

c = 25.1 J °C-1mol-1 n = (0.95 x 103 g)/(55.85 g mol-1) = 17.01 mol

T = 125.0°C - 20.0°C = 105.0°C

Plugging these values into the equation then gives:

Heat = ( 25.1 J °C-1mol-1)(17.01 mol)(105.0°C) = 4.5 x 104 J

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3. What is the heat capacity of a calorimeter that contains 125.0 g of water, if it took 40.0 kJ to raise the temperature of the calorimeter and the water 14.00 °C? (Assume a constant heat capacity for water of 4.184 J g-1°C-1)

In this problem we are being asked to determine how much heat it takes to change the temperature of an instrument (the calorimeter). We are given the heat to change the temerature of the instrument AND the water it contains.

We can solve this by first finding the heat capacity of the apparatus + water, and then subtract the heat capacity of the water it contains.

Total heat to raise temperature 14.00°C= 40.0 kJ = 40.0 x 103 J

Heat to raise 125.0 g water 14.00°C = (14.00°C)(4.184 J g-1°C-1)(125.0 g) = 7.3238 x 103 J

Heat to raise temperature of calorimeter 14.00 °C = 40.0 x 103 J - 7.3238 x 103 J = 3.2676 x 104 J

Heat capacity of calorimeter = (3.2676 x 104 J)/(14.00 °C) = 2.334 x 103 J °C-1 = 2.33 x 103 J °C-1

 

4. 4. A 75.0 g bar of copper metal at 128.2 °C is dropped into 1.000 L of cool water in an insulated container where its temperature drops to 24.1 °C, in equilibrium with the water. Assuming all of the heat goes to heat up the water (the container has a heat capacity of zero), what was the initial temperature of the water? (The molar heat capacity of copper is 24.4 Jmol-

1°C-1, the heat capacity of water is 4.184 Jg-1°C-1.)

We can begin by finding how much heat was transferred to the water from the copper:

Heat = cn T

To solve this we need:

molar heat capacity (n appears in the equation), c = 24.4 Jmol-1°C-1 moles of copper, n = (75.0g/63.55gmol-1) = 1.180 mol

temperature change, T = 128.2 - 24.1°C = 104.1°C

Heat = (24.4 Jmol-1°C-1)(1.180 mol)(104.1°C) = 2,997 J

To find the change in temperature of the water can then be found using dimensional analysis and the heat capacity of water. Looking at the heat capacity:

4.184 Jg-1°C-1

We see that it will have to be in the bottom with the mass of water (1.000 x 103mL*1.00g/mL = 1.000 x 103g) and the energy on top:

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T = (2,997 J)/(4.184 Jg-1°C-1)(1.000 x 103g) = 0.7164°C

Finally then, the original water temperature, To must have been:

To = Tf - T

To = 24.1 - 0.7164°C = 23.38°C = 23.4°C

 

5. A 0.828 g sample of methanol is completely combusted in a bomb calorimeter with excess oxygen. The calorimeter contains 1.35 kg of water and the instrument has a heat capacity of 1.06 kJ °C-1 without water. Combustion of the methanol increases the temperature of the calorimeter increases from 23.10 to 25.90°C. Find the molar heat of combustion of methanol (heat released in burning one mole of methanol).

In this situation the heat is absorbed by both the instrument and the water it contains, so we can break this portion of the problem into two parts, the heat absrbed by the water, and the heat absorbed by the instrument.

i. Heat absorbed by water:

Heatwater = cm T, where m = mass in grams

Heatwater =(4.184 Jg-1°C-1)(1.35 x 103g)(25.90-23.10°C) = 1.5816 x 104J

ii. Heat absorbed by instrument:

Heatinst. = cinst. T

Heatinst. = (1.06 kJ °C-1)(25.90-23.10°C) = 2.968 x 103J = 0.2968 x 104J

iii. Total heat:

Heattot. = Heatwater + Heatinst. = 1.5816 x 104J + 0.2968 x 104J = 1.8784 x 104J

We now need to find the number of moles of methanol which provided this heat:

Moles = 0.828g/32.0gmol-1 = 2.59 x 10-2 mol

Finally, the molar heat of methanol is heat/mol:

(1.8784 x 104J)/(2.59 x 10-2 mol) = 7.26 x 105J mol-1

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