Temperature and Thermal Energy of coffee cup

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SUMMARY

The discussion focuses on calculating the heat absorbed by a glass coffee cup when it is immersed in hot dishwater. The specific heat formula used is Q = mC∆T, where m is the mass of the cup, C is the specific heat capacity, and ∆T is the change in temperature. The mass of the cup is converted from grams to kilograms, resulting in a final calculation of Q = 100,320 J. This demonstrates the application of thermodynamic principles in practical scenarios.

PREREQUISITES
  • Understanding of specific heat capacity (C)
  • Knowledge of mass conversion from grams to kilograms
  • Familiarity with temperature change calculations (∆T)
  • Basic grasp of thermodynamics principles
NEXT STEPS
  • Study the concept of specific heat capacity in detail
  • Learn about the laws of thermodynamics and their applications
  • Practice problems involving heat transfer calculations
  • Explore the differences between thermal energy and temperature
USEFUL FOR

Students preparing for physics courses, educators teaching thermodynamics, and anyone interested in practical applications of heat transfer principles.

mppaki
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Before I begin, Please note that for some reason I don't really grasp physics equations which is odd since I had a A in chemistry but physics just confuses me and also, I am taking this class to prepare me for college. My pencil broke and I didn't think we would have homework of the stuff we learned so quick, so I decided not to ask anyone for a pencil and just listened through the rest of the lecture but now I regret it.

1. A 4.00 x 10²-g glass coffee cup is at room temperature, 20.0°C. It is then plunged into hot dishwater, 80.0°C. If the temperature of the cup reaches that of the dishwater, how much heat does the cup absorb? Assume the mass of the dishwater is large enough so its temperature doesn't change appreciably.



Homework Equations


The only thing I can think of is Q=mC∆T, but I am pretty sure that is the wrong formula.


The Attempt at a Solution



m = 4.00 x 10²g = 4000 / 1000 = 4kg.
Ti = 20.0°C
Tf = 80.0°C
C = 4180

Q = (4)(4180)(80-20)
Q = 1,003,200 j/kg * K
 
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m = 4.00 x 10²g = 4000 / 1000 = 4kg.
Check this one.
 
oh, um the 1000 is because i needed to convert the grams to kilograms.
 
It should be 400/1000 kg
 
oh yes lol, I have a habit of entering in my calc the x10...

Q = (0.4)(4180)(80-20)
Q = 100,320 J/Kg * k
 

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