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Ch.5 - Thermochemistry
Chapter 5, Problem 105

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately 1.8 kg. The specific heat of the brick is 0.85 J/g•K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7⨉103 gal of water?

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1
Calculate the total mass of water in kilograms, knowing that 1 gallon of water weighs approximately 3.785 kg.
Convert the mass of water from kilograms to grams, since the specific heat is given in J/g•K.
Calculate the total heat capacity of the water using the formula: Heat Capacity = mass (g) × specific heat (J/g•K) × temperature change (ΔT). Since we are comparing capacities, ΔT can be considered as 1 K for simplification.
Calculate the heat capacity of one brick using its mass (1.8 kg converted to grams) and its specific heat (0.85 J/g•K) with ΔT as 1 K.
Divide the total heat capacity of the water by the heat capacity of one brick to find the number of bricks needed to match the heat capacity of the water.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin). It is a crucial property that determines how much energy a material can store. In this question, the specific heat of the bricks (0.85 J/g•K) indicates how effectively they can absorb and retain heat compared to water.
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Heat Capacity

Heat Capacity

Heat capacity is the total amount of heat energy a substance can store, which depends on both its mass and specific heat capacity. It is calculated by multiplying the mass of the substance by its specific heat. In this scenario, we need to equate the heat capacity of the bricks to that of a large volume of water to determine how many bricks are needed.
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Conversion of Units

Conversion of units is essential in chemistry to ensure that measurements are compatible when performing calculations. In this question, the volume of water (1.7×10^3 gallons) must be converted to mass (in grams) to compare it with the mass of the bricks. Understanding how to convert gallons to liters and then to grams using the density of water is necessary for solving the problem.
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Related Practice
Textbook Question

A sample of gas is contained in a cylinder-and-piston arrangement. There is an external pressure of 100 kPa. The gas undergoes the change in state shown in the drawing. (b) Now assume that the cylinder and piston are made up of a thermal conductor such as a metal. During the state change, the cylinder gets colder to the touch. What is the sign of q for the state change in this case? Describe the difference in the state of the system at the end of the process in the two cases. What can you say about the relative values of E?

Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (a) Determine the amount of heat, in J, lost by the copper block.

Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (b) Determine the amount of heat gained by the water. The specific heat of water is 4.184 J/1gK.

Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?