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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Hey everyone, we're told that a specific type of coolant has a specific heat of 0.963 jewels per grams times kelvin, which is actually lower than that of water. However, in terms of practicality, coolant can be better in certain situations. One property among others is having a much lower freezing point and a much higher boiling point than water, which makes it more robust and has a higher temperature range for operating as a liquid. Considering that the coolant has a density of 1.79 g per millimeter, calculate the equivalent volume in gallons needed to provide the same heat capacity as five gallons first. Let's go ahead and calculate the heat capacity of five gallons of water and we can do this by taking the specific heat of water which is 4.184 joules per gram times kelvin And multiplying it by five gallons. Now essentially we want to use our dimensional analysis to get the heat capacity of our five gallon water and we know that our units for heat capacity is going to be jewels per kelvin. So essentially we want to use our dimensional analysis in order to get rid of the other units. So we know that we have 3.785 liters per one gallon and we know that one leader is equivalent to 10 to the third milliliters. And now we can go ahead and use the density of water which is one g per mil leader. Now, when we calculate this out and cancel out all of our units, we find that we end up with the heat capacity of 7.918 times 10 to the four joules per kelvin. Now we want to go ahead and use the heat capacity of our five gallon water in order to calculate the volume of our coolant. So we're going to go ahead and take that value of 7.918 times 10 to the four joules per kelvin. And we want to divide this by the specific heat of our coolant, which was 0.963 jewels over grams times kelvin. Next, we want to use dimensional analysis to end up with gallons for our volume of coolant. So taking the density of our coolant, We know that we have 1.79 g per one mill leader. And again we know that we have 10 to the 3rd ml per one leader, And we know that 3.785 L is equivalent to one gallon. So when we calculate this out and cancel out our units, We end up with a volume of 12.1 gallons. And this is going to be our final answer. Now I hope that made sense. And let us know if you have any questions

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Key Concepts

Specific Heat Capacity

Heat Capacity

Conversion of Units