LEAD IS A SOFT, DENSE METAL WITH A SPECIFIC HEAT OF 0.028KCAL/KGC, A MELTING POINT OF 328.0C, AND A HEAT FUSION OF5.5KCAL/KG. HOW MUCH HEAT MUST BE PROVIDED TO MELT A 250.0KG SAMPLE OF LEAD WITH A TEMPERATURE OF 20.0C?
Figure the heat to heat if from 20C to the melting point (Q=mc deltaT)
Add the heat needed to melt it at the melting point (Q=heatfusion*mass)
I CAME UP WITH 1,640KCAL I AM RIGHT?
No. Break it down in to calculations as I laid out, and let me see them.
A heat pump releases 60.0 Kcal as it removes 40.0 Kcal at the evaporate coils. How much work does this heat pump ideally accomplish
Sure, here's how you can calculate the heat required to melt the lead sample:
1. First, calculate the heat required to raise the temperature of the lead sample from 20.0°C to its melting point of 328.0°C using the formula Q = mcΔT, where:
- Q is the heat required
- m is the mass of the lead sample (250.0 kg)
- c is the specific heat of lead (0.028 kcal/kg°C)
- ΔT is the change in temperature (melting point - initial temperature = 328.0 - 20.0 = 308.0°C)
Substituting the given values into the formula, we get:
Q = (250.0 kg) * (0.028 kcal/kg°C) * (308.0°C)
Q = 2156 kcal
2. Next, calculate the heat required to actually melt the lead sample at its melting point using the formula Q = heat fusion * mass, where:
- Q is the heat required
- heat fusion is the heat of fusion for lead (5.5 kcal/kg)
- mass is the mass of the lead sample (250.0 kg)
Substituting the given values into the formula, we get:
Q = (5.5 kcal/kg) * (250.0 kg)
Q = 1375 kcal
Finally, add the heat required to raise the temperature to the melting point and the heat required for actual melting:
Total heat required = heat to raise temperature + heat for melting
Total heat required = 2156 kcal + 1375 kcal
Total heat required = 3531 kcal
So, the correct answer is that 3531 kcal of heat must be provided to melt the 250.0 kg sample of lead at an initial temperature of 20.0°C.
To calculate the heat required to raise the temperature of lead from 20.0°C to its melting point at 328.0°C, we can use the equation Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
m (mass) = 250.0 kg
c (specific heat) = 0.028 kcal/kg°C
ΔT (change in temperature) = 328.0°C - 20.0°C = 308.0°C
Calculating Q for heating:
Q = mcΔT
= (250.0 kg)(0.028 kcal/kg°C)(308.0°C)
= 2156 kcal
The heat required to raise the temperature of the lead to its melting point is 2156 kcal.
Next, we need to calculate the heat required to melt the lead at its melting point. We can use the equation Q = heat fusion * mass, where Q is the heat, heat fusion is the heat of fusion, and mass is the mass of the sample.
Given:
heat fusion = 5.5 kcal/kg
mass = 250.0 kg
Calculating Q for melting:
Q = heat fusion * mass
= (5.5 kcal/kg)(250.0 kg)
= 1375 kcal
The heat required to melt the lead at its melting point is 1375 kcal.
To find the total heat required, we need to add the heat required for heating and the heat required for melting:
Total heat required = heat for heating + heat for melting
= 2156 kcal + 1375 kcal
= 3531 kcal
Therefore, the correct answer is 3531 kcal, not 1640 kcal.
To calculate the total heat required to melt a 250.0 kg sample of lead with a temperature of 20.0°C, follow these steps:
1. Calculate the heat required to heat the lead from 20.0°C to its melting point (328.0°C).
To do this, use the equation Q = mcΔT, where:
Q is the heat (in kcal)
m is the mass of the lead sample (in kg)
c is the specific heat of lead (in kcal/kg°C)
ΔT is the temperature change (in °C)
So, for this step:
m = 250.0 kg
c = 0.028 kcal/kg°C
ΔT = (melting point - starting temperature) = (328.0°C - 20.0°C) = 308.0°C
Q1 = (250.0 kg) * (0.028 kcal/kg°C) * (308.0°C) = 2156 kcal
2. Calculate the heat required to melt the lead at its melting point.
To do this, use the equation Q = heat fusion * mass, where:
Q is the heat (in kcal)
heat fusion is the heat of fusion of lead (in kcal/kg)
mass is the mass of the lead sample (in kg)
So, for this step:
heat fusion = 5.5 kcal/kg
mass = 250.0 kg
Q2 = (5.5 kcal/kg) * (250.0 kg) = 1375 kcal
3. Add the heat calculated in step 1 and step 2 to find the total heat required.
Total heat = Q1 + Q2 = 2156 kcal + 1375 kcal = 3531 kcal
Therefore, the correct answer is 3531 kcal, not 1640 kcal.