How much heat is absorbed by the calorimeter if the temperature change is 12 C and the calorimeter constant (Heat Capacity) is 3.3 J/K?
Tennessee State University is NOT the name of your school subject.
I answered this question when you posted it earlier. Go back and look at your posts to see if you have a response before you post again, please.
How much heat is absorbed by the calorimeter if the temperature change is 11 C and the calorimeter constant (Heat Capacity) is 3.3 J/K?
To determine the amount of heat absorbed by the calorimeter, you can use the formula:
Q = C * ΔT
Where:
Q is the heat absorbed by the calorimeter,
C is the calorimeter constant (also known as heat capacity), and
ΔT is the change in temperature.
In this case, the temperature change (ΔT) is 12°C, and the calorimeter constant is 3.3 J/K.
Plugging in the values, we have:
Q = 3.3 J/K * 12°C
Now, we need to convert the temperature change from Celsius to Kelvin, as the units for temperature in the formula are in Kelvin.
To convert from Celsius to Kelvin, we use the simple formula: K = °C + 273.
ΔT(K) = 12°C + 273 = 285 K
Now we can substitute the value of ΔT in Kelvin into the formula:
Q = 3.3 J/K * 285 K
Calculating this, we find:
Q = 940.5 J
Therefore, approximately 940.5 Joules of heat are absorbed by the calorimeter.