Final temperature after calories are added

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SUMMARY

The discussion centers on calculating the final temperature of a 60.5g piece of lead after adding 328 calories of heat. The relevant equation is Q = mCΔT, where Q represents heat added, m is mass, C is specific heat capacity, and ΔT is the change in temperature. Participants emphasized the necessity of using consistent units and correctly identifying the specific heat capacity of lead, which is approximately 0.128 cal/g°C. The final temperature can be determined by rearranging the equation to solve for the final temperature after calculating ΔT.

PREREQUISITES
  • Understanding of the specific heat capacity concept
  • Familiarity with the equation Q = mCΔT
  • Basic knowledge of unit conversion in thermal calculations
  • Experience with solving algebraic equations
NEXT STEPS
  • Research the specific heat capacity of various materials, focusing on metals
  • Learn about unit conversion techniques in thermal physics
  • Explore examples of heat transfer calculations in different contexts
  • Practice solving problems involving heat transfer and temperature changes
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Students studying thermodynamics, physics educators, and anyone interested in heat transfer calculations in materials.

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Homework Statement



A 60.5g piece of lead is at 18.6oC. If 328 cal of heat is added to the lead, what is its final temperature? Do not enter units.

Homework Equations


I was trying to use 328 as the Q value and solve for delta T
Q=mCdelta T
HOWEVER, this wasn't working.



The Attempt at a Solution


Could someone please explain to me what equation I need to use and how the 328 cal are used? Thanks.
 
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Can you show your calculation? You say you solved for deltaT but you need to go one step further and find the final temperature. If you did that, are you making sure your units are consistent? What are you using for C?
 

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