Specific heat : solids and liquids

AI Thread Summary
The discussion revolves around solving a homework problem related to the specific heat of lead. The specific heat of lead is given as 31.0 calories per kilogram per Celsius degree, and the task is to calculate the heat required to raise the temperature of 1.30 kg of lead from 23.0 degrees Celsius to its melting point of 327.3 degrees Celsius, as well as the heat needed for the phase change from solid to liquid using the heat of fusion. Participants emphasize the importance of using the correct equations, specifically mCΔT for temperature change and the heat of fusion for phase changes. The original poster expresses initial confusion but realizes that the problems become manageable with the right approach. Understanding these concepts is crucial for solving similar problems effectively.
airforce840
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I am having a lot of trouble with these types of problems.. this one is on a homework assignment and i don't know where to begin.
heres the problem :
The sepcific heat of lead is 31.0 calories per kilogram per Celsius degree. (a) Find the heat necessary to raise the temperature of 1.30 kg of lead from 23.0 degrees Celsius to its melting point, 327.3 C. (b) Find the heat necessary for the phase change from solid to liquid. The heat of fusion for lead is 24.5 x 10^3 joules per kilogram.
Any help is great.
Thanks in advanced
Patrick
 
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airforce840 said:
The sepcific heat of lead is 31.0 calories per kilogram per Celsius degree.
(a) Find the heat necessary to raise the temperature of 1.30 kg of lead from 23.0 degrees Celsius to its melting point, 327.3 C.

What equation do you have that relates heat, specific heat, and temperature change?

(b) Find the heat necessary for the phase change from solid to liquid. The heat of fusion for lead is 24.5 x 10^3 joules per kilogram.

What equation do you have that relates heat, heat of fusion (aka latent heat of fusion) and mass?
 
i feel so stupid.. these problems are quite easy once i looked at it again.. all i got to do is use m C delta T and I am good 2 go.
 
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