lead melts at 328 degrees C. lead has a heat of fusion of 2.04X10^4 J/kg and a specific heat of 130 J/kgK. how much heat must be removed from 0.50 kg of molten lead (liquid) at 328 degrees C to make it all a solid chunk at 300 degrees C?
heat= Hf*mass+mass*specific heat*(328-300)
To calculate the heat that must be removed, we need to consider two steps: first, solidifying the molten lead at 328 degrees C to a solid state at 327.5 degrees C, and then cooling it further to 300 degrees C.
Step 1: Solidifying the lead from 328 degrees C to 327.5 degrees C.
To calculate the heat required to solidify the lead, we use the formula:
Q = m * ΔHf
where:
Q is the heat required (in joules),
m is the mass of the lead (in kilograms), and
ΔHf is the heat of fusion (in joules per kilogram).
Given:
m = 0.50 kg
ΔHf = 2.04 × 10^4 J/kg
Substituting the values into the formula, we get:
Q1 = (0.50 kg) * (2.04 × 10^4 J/kg)
Q1 = 1.02 × 10^4 J
Step 2: Cooling the solid lead from 327.5 degrees C to 300 degrees C.
To calculate the heat required to cool the solid lead, we use the formula:
Q = m * c * ΔT
where:
Q is the heat required (in joules),
m is the mass of the lead (in kilograms),
c is the specific heat capacity of lead (in joules per kilogram per degree Celsius), and
ΔT is the temperature change (in degrees Celsius).
Given:
m = 0.50 kg
c = 130 J/kgK
ΔT = 327.5 - 300 = 27.5 degrees C
Substituting the values into the formula, we get:
Q2 = (0.50 kg) * (130 J/kgK) * (27.5 degrees C)
Q2 = 1.79 × 10^3 J
Total heat required = Q1 + Q2
= 1.02 × 10^4 J + 1.79 × 10^3 J
= 1.20 × 10^4 J
Therefore, approximately 1.20 × 10^4 J of heat must be removed from 0.50 kg of molten lead at 328 degrees C to make it a solid chunk at 300 degrees C.