How many joules of heat are lost by 3399kg granite as it cools from 41.7 degrees Celcius to -13.3 degrees Celcius? answer must be in joules not kJ
q = mass granite x specific heat granit x delta T.
ok so,
q=3399kg x 0.803 x 55
q = 150.116
Is that correct?
so then my answer would be 150.116kJ which equals 150116835J. Is that correct???
To calculate the amount of heat lost by the granite, we can use the specific heat capacity equation:
Q = m * c * ΔT
where:
Q is the heat energy lost (in joules),
m is the mass of the granite (in kilograms),
c is the specific heat capacity of granite (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = (-13.3 °C) - (41.7 °C)
ΔT = -55 °C
Next, we need to find the specific heat capacity of granite. The specific heat capacity represents the amount of heat energy required to raise the temperature of an object by 1 degree Celsius per unit mass.
The specific heat capacity of granite is typically around 0.79 J/(g°C), but we need to convert it to joules per kilogram per degree Celsius:
1 g = 0.001 kg
0.79 J/(g°C) = 0.79 J/(0.001 kg°C)
0.79 J/(g°C) = 790 J/(kg°C)
Now, we can substitute the values into the specific heat capacity equation:
Q = m * c * ΔT
Q = (3399 kg) * (790 J/(kg°C)) * (-55 °C)
Q = - 136,256,550 J
Therefore, the granite loses 136,256,550 joules of heat energy as it cools from 41.7 degrees Celsius to -13.3 degrees Celsius.