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Transcript
q=CmΔT, as easy as 1, 2, 3!
The End :)
Another Practice Problem
A 90.4 piece of Cu is heated from 4 degrees Celsius to 76 degrees Celsius. Specific heat is .389 j/g. How much heat is needed?
Practice Problems:
Temperature of 95.4g Cu increases from 25.0 degrees Celsius to 48 degrees Celsius when Cu absorbs 849 joules of heat. What is the specific heat?
Answer
q= .389*90.4*72
What do the letters mean?
q=2.53192
2.53 kj
q= heat (joules)
C= specific heat (j/g*c)
m= mass (g)
ΔT= temperature change
THe Answer
849=C*95.4*48
849=C*4579.2
849/4579.2=
What this formula is used For:
.387 j/g*°C
q=CmΔT can be used to solve for calories, or joules. Or any of its other parts, so you can use it to solve for specific heat (C), mass (m) and temperature change (ΔT).
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