A solar home contains 3.50 multiplied by 105 kg of concrete (specific heat = 1.00 kJ/kg·K). How much heat is given off by the concrete when it cools from 29 to 20°C?
To calculate the amount of heat given off by the concrete, you can use the formula:
Q = m * c * ΔT
where:
Q is the amount of heat (in joules),
m is the mass of the concrete (in kilograms),
c is the specific heat capacity of concrete (in kJ/kg·K),
ΔT is the change in temperature (in degrees Celsius).
Given:
m = 3.50 * 10^5 kg
c = 1.00 kJ/kg·K
ΔT = 29°C - 20°C = 9°C
First, convert the units of c from kJ/kg·K to J/kg·K:
1 kJ = 1000 J
c = 1.00 kJ/kg·K = 1.00 * 1000 J/kg·K = 1000 J/kg·K
Now you can substitute the given values into the formula and solve for Q:
Q = (3.50 * 10^5 kg) * (1000 J/kg·K) * (9°C)
Q = 3.50 * 10^5 * 1000 * 9 J
Q = 31.5 * 10^8 J
Therefore, the concrete gives off 31.5 * 10^8 J of heat when it cools from 29 to 20°C.
To calculate the heat given off by the concrete, we need to use the specific heat formula:
Heat = mass * specific heat * change in temperature
Here's how we can use this formula to find the answer to your question:
1. Determine the mass of the concrete:
Given that the solar home contains 3.50 * 10^5 kg of concrete, the mass of the concrete is 3.50 * 10^5 kg.
2. Determine the change in temperature:
The concrete cools from 29 to 20°C. Therefore, the change in temperature is 29°C - 20°C = 9°C.
3. Determine the specific heat of concrete:
The specific heat of concrete is given as 1.00 kJ/kg·K.
Now, let's plug in these values into the formula:
Heat = mass * specific heat * change in temperature
= (3.50 * 10^5 kg) * (1.00 kJ/kg·K) * (9°C)
First, convert the mass from kg to g and the change in temperature from °C to K:
3.50 * 10^5 kg = 3.50 * 10^8 g
9°C = 9 K
Now calculate the heat:
Heat = (3.5 * 10^8 g) * (1.00 kJ/g·K) * (9 K)
= 31.5 * 10^8 kJ
= 3.15 * 10^9 kJ
Therefore, the concrete gives off 3.15 * 10^9 kJ of heat when it cools from 29 to 20°C.