Temperature and Thermal Energy of coffee cup

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Homework Help Overview

The discussion revolves around a physics problem involving the heat absorbed by a glass coffee cup when placed in hot dishwater. The subject area includes concepts of thermal energy and temperature change.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the heat transfer equation Q=mC∆T, with some expressing uncertainty about its correctness. There are attempts to clarify the mass conversion from grams to kilograms, and questions arise regarding the calculations involved.

Discussion Status

Participants are actively engaging with the problem, checking each other's calculations and clarifying the mass conversion. There is no explicit consensus on the final calculation, but there is a collaborative effort to understand the equation and its application.

Contextual Notes

Some participants mention confusion regarding physics equations compared to their experience in chemistry, indicating a potential struggle with the material. Additionally, there is a reference to a lack of preparation due to unexpected homework assignments.

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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