How much heat (in kJ) is given when 85.0 grams of lead cools from 200.0 degrees Celsius to 10.0 degrees Celsius? (Cp of lead = 0.129 J/g)
q = mass Pb x specific heat Pb x (Tfinal-T(initial)
-2.08
To find the amount of heat given when a substance cools, you need to use the equation:
q = m × Cp × ΔT
where:
q is the heat transferred,
m is the mass of the substance,
Cp is the specific heat capacity of the substance,
and ΔT is the change in temperature.
In this case, you are given:
m = 85.0 grams (the mass of lead),
Cp = 0.129 J/g (the specific heat of lead),
ΔT = (final temperature) - (initial temperature) = 10.0°C - 200.0°C = -190.0°C.
However, before we can proceed with the calculation, we need to convert the mass from grams to kilograms since the specific heat capacity is in joules per gram.
Converting grams to kilograms:
85.0 grams = 85.0 grams × (1 kilogram / 1000 grams) = 0.085 kilograms
Now we can calculate the heat transferred (q):
q = m × Cp × ΔT
= 0.085 kg × 0.129 J/g × -190.0°C
= 0.085 kg × 0.129 J/g × -190.0°C
= -24.579 J
Since we want the answer in kJ (kilojoules), we can convert the heat transferred from joules to kilojoules by dividing by 1000:
q = -24.579 J ÷ 1000
= -0.0246 kJ
Therefore, the amount of heat given when 85.0 grams of lead cools from 200.0°C to 10.0°C is approximately -0.0246 kJ. Note that the negative sign indicates that heat is being lost.