Answer:
Q = - 8501.99 j
Explanation:
Given data:
Specific heat of Al = 0.902 j/g.°C
Heat lost = ?
Mass of sample = 23.984 g
Initial temperature = 415°C
Final temperature = 22°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 22°C - 415°C
ΔT = -393°C
Q = m.c. ΔT
Q = 23.984 g× 0.902 j/g.°C × -393°C
Q = - 8501.99 j
The quantity of heat lost when the piece of aluminum cools from 415°C to 22°C is -8501.99 Joules.
Given the following data:
To find how much heat is lost when the piece of aluminum cools:
Mathematically, heat capacity or quantity of heat is given by the formula;
[tex]Q = mc\theta[/tex]
Where:
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity.
∅ represents the change in temperature.
[tex]\theta = T_2 - T_1\\\\\theta = 22 - 415\\\\\theta = -393[/tex]
Substituting the values into the formula, we have:
[tex]Q = 23.984(0.902)(393)[/tex]
Q = -8501.99 Joules.
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