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?
Tell me if this is right
For the first step this is what I am doing
m=250.0kg
c=0.028kcal
deltaT- 328-20?? Is this correct
this gives me 2156
Am I on the right track for the first step?
Yes, but your number doesn't have units. Most profs will count off for that.
I know.. I am just making sure I am on the right track.. I am going to try and finish it now.. do you mind if I send you my answer and you tell me how I did?
That looks ok to me.
To find the amount of heat required to melt a sample of lead, we need to consider two factors: the heat required to raise the temperature of the lead from 20.0°C to its melting point, and the heat required to melt the lead at its melting point.
First, let's calculate the heat required to raise the temperature of the lead from 20.0°C to its melting point:
Heat = mass x specific heat x change in temperature
mass = 250.0 kg
specific heat = 0.028 kcal/kg°C
change in temperature = 328.0°C - 20.0°C
Heat = 250.0 kg x 0.028 kcal/kg°C x (328.0°C - 20.0°C)
Next, let's calculate the heat required to melt the lead at its melting point:
Heat = mass x heat fusion
mass = 250.0 kg
heat fusion = 5.5 kcal/kg
Heat = 250.0 kg x 5.5 kcal/kg
Finally, we can add the two calculated heats together to get the total heat required to melt the lead:
Total Heat = Heat to raise the temperature + Heat to melt
Total Heat = (250.0 kg x 0.028 kcal/kg°C x (328.0°C - 20.0°C)) + (250.0 kg x 5.5 kcal/kg)
Now, you can calculate the value for the total heat required to melt the 250.0 kg sample of lead.
q1 = heat to move temperature from 20 C to 328.
q1 = mass Pb x specific heat lead x delta T.
q2 = heat to melt Pb at 328.
q2 = mass Pb x heat fusion.
Total heat = q1 + q2.
Watch the units. Post your work if you get stuck.
okay is this correct
I took 250.0 * 5.5 = 1375
1375+2156=3531kcal