A block of aluminum weighing 140. g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calculate the specific heat of aluminum.
The specific heat (c) of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius.
To calculate the specific heat of aluminum, we can use the formula:
Q = m * c * ΔT
Where:
- Q is the heat energy released (1080 J)
- m is the mass of the aluminum (140 g)
- c is the specific heat of aluminum (unknown)
- ΔT is the change in temperature (98.4°C - 62.2°C = 36.2°C)
Plugging in the values into the formula, we get:
1080 J = 140 g * c * 36.2°C
Simplifying the equation:
c = 1080 J / (140 g * 36.2°C)
c = 1080 J / 5056 g°C
c ≈ 0.213 J/g°C
Therefore, the specific heat of aluminum is approximately 0.213 J/g°C.
To calculate the specific heat of aluminum, we can use the formula:
q = m * c * ΔT
Where:
- q is the heat released in joules,
- m is the mass of the aluminum in grams,
- c is the specific heat capacity of aluminum, and
- ΔT is the change in temperature in Celsius.
Given:
- The mass of aluminum, m = 140 g
- The change in temperature, ΔT = (62.2°C - 98.4°C) = -36.2°C (negative because the aluminum is being cooled)
- The heat released, q = -1080 J (negative because heat is being released)
Substituting the given values into the formula, we get:
-1080 J = 140 g * c * (-36.2°C)
To solve for c, we rearrange the equation:
c = -1080 J / (140 g * -36.2°C)
c = 1080 J / (140 g * 36.2°C)
c ≈ 0.215 J/g°C
Therefore, the specific heat of aluminum is approximately 0.215 J/g°C.
To calculate the specific heat of aluminum, we can use the equation:
q = m * c * ΔT
where:
- q is the heat absorbed or released by the substance (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
Given data:
- q = -1080 J (negative because heat is released)
- m = 140 g
- ΔT = (final temperature) - (initial temperature) = 62.2°C - 98.4°C = -36.2°C (Note: we use a negative value since the temperature decreases)
Substituting the given values into the equation, we have:
-1080 J = 140 g * c * (-36.2°C)
Now, let's solve for c:
c = -1080 J / (140 g * (-36.2°C))
Calculating the above expression gives us:
c ≈ 0.385 J/g°C
Therefore, the specific heat of aluminum is approximately 0.385 J/g°C.