Question

my home has a passive solar energy features. The interior of the house has brickwork incorporated in as a heat absorber. Each brick wighs approximatelly 1.65 kg. The specific heat ot the brick is 0.76 J/g*C. How many brick do I have incorporated into the interior of my house to provide the same total molar heat capacity of 2.3x10^3 gallons of water? Specific heat of water is 40184 J/G*c

Answer 1

Heat absorbed by 2.3 times 10^3 gallons of water = 40184 J/g degree c times 1000 times 5.456 times 2.3 times 10^3 = 4.2 times 10^11 J/degree C grams No. of bricks = 4.2 times 10^11/0.76 = 5.526 times 10^11 grams. No. of bricks = 5.526 times 10^22/1.63 times 10^3 = 3.35 times 10^8