The temperature of a 2.0 g sample of aluminum increases from 25 to 30 degrees celcius. How many joules of heat energy were added? Specific heat of Aluminum is .9
9.0 J
Well, well, well, looks like we have a hot topic here! To calculate the heat energy added, we can use the formula:
Q = m * c * ∆T
Where Q is the heat energy, m is the mass, c is the specific heat, and ∆T is the change in temperature. So let's plug in the numbers, shall we?
Q = 2.0 g * 0.9 J/g°C * (30°C - 25°C)
Now, let's do some math magic:
Q = 2.0 g * 0.9 J/g°C * 5°C
Q = 9 J
Voila! **9 joules** of heat energy were added. That's enough to warm the heart of any aluminum sample!
To calculate the heat energy added to the sample, we can use the formula:
Q = m * c * ΔT
where:
Q is the heat energy added,
m is the mass of the sample,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
Given:
m = 2.0 g (mass of the aluminum sample),
c = 0.9 J/g·°C (specific heat capacity of aluminum),
ΔT = 30°C - 25°C = 5°C (change in temperature).
Plugging these values into the equation:
Q = 2.0 g * 0.9 J/g·°C * 5°C
Calculating:
Q = 9 J
Therefore, 9 joules of heat energy were added to the aluminum sample.
To calculate the heat energy added to the aluminum sample, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy added (in joules),
m is the mass of the aluminum sample (in grams),
c is the specific heat of aluminum (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
Given values:
m = 2.0 g (mass of sample)
c = 0.9 J/g°C (specific heat of aluminum)
ΔT = (final temperature) - (initial temperature)
= 30°C - 25°C
= 5°C
Substituting the values into the formula:
Q = (2.0 g) * (0.9 J/g°C) * (5°C)
Q = 9 Joules
Therefore, 9 joules of heat energy were added to the aluminum sample.