Aluminum has a specific heat of 0.902 Jg-1K-1. How much heat is lost when a piece ofaluminum with a mass of 23.984 g cools from a temperature of 415.0° C to atemperature of 22.0° C?
Question
Aluminum has a specific heat of 0.902 Jg-1K-1. How much heat is lost when a piece ofaluminum with a mass of 23.984 g cools from a temperature of 415.0° C to atemperature of 22.0° C?
Solution
To solve this problem, we will use the formula for heat transfer:
q = mcΔT
where:
- q is the heat transferred,
- m is the mass of the substance,
- c is the specific heat capacity of the substance, and
- ΔT is the change in temperature.
Given:
- m = 23.984 g,
- c = 0.902 J/g·K,
- Initial temperature (T_initial) = 415.0°C,
- Final temperature (T_final) = 22.0°C.
First, we need to convert the temperatures from Celsius to Kelvin. However, since we are interested in the change in temperature (ΔT), and because the size of the degree is the same in Celsius and Kelvin, we can use the temperatures in Celsius.
So, ΔT = T_final - T_initial = 22.0°C - 415.0°C = -393.0°C.
Now, we can substitute the given values into the formula:
q = mcΔT q = (23.984 g)(0.902 J/g·K)(-393.0°C) q = -8522.6 Joules
The negative sign indicates that the heat is lost. So, the piece of aluminum loses 8522.6 Joules of heat when it cools from 415.0°C to 22.0°C.
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