Energy Transfer in Different Scenarios

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The discussion revolves around calculating energy transfer in various scenarios, including cooling coffee, sweating during volleyball practice, heating water, and melting ice. Key equations involve heat transfer formulas, specifically Q = m*c*ΔT, where m is mass, c is heat capacity, and ΔT is the change in temperature. Participants seek clarification on how to apply these concepts to specific problems, such as determining energy released by cooling coffee or absorbed by evaporating sweat. The importance of understanding heat balance, where heat gained equals heat lost, is emphasized throughout the conversation. Overall, the thread serves as a collaborative effort to solve energy transfer calculations in different contexts.
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Homework Statement


a cup of coffee (140g) cools from 75 degrees C down to 20. degrees C. how much energy does it release to the surroundings? answer in J

suppose dutring volleyball practice, you lost 2.0 lbs of water due to sweating. if all of this water evaporated, how much energy did the water absorb from your body? answer in kJ.

suppose during a lab that 65kJ of energy was transferred to 450 g of water at 20C. what would be the final temperature of the water? answer in C

the heat capacity of solid iron is 0.447 J/g degrees C. If 65kJ was transferred to a 450 g chunk of ioron at 20. C what would be the final temperature?

a serving of cheezits releases 130kcal when digested by your body. if this same amount of energy was transferred to 2.5kg of water at 27 C what would the final temperature be?

if 130kcal was transferred to 2.5kg of water at its boiling point, what fraction of the water would be vaporized?

suppose a bag full of ice (450g) at 0.0 C sits on the counter and begins to melt to a liquid state. how much energy must be absorbed by the ice if 2/3 of it melted?

Homework Equations



Energy constants
334 J/g heat of fusion
2260 J/g heat of vaporization
2.1 J/g heat capacity of solid water
4.18 J/g heat capacity of liquid water

The Attempt at a Solution

If you could show work with answer that would be very apreciated as i wasnt there the day when he lectured us on how to do this. an explanation of the problems would be awesome too!

i assume that you multiply the mass times the heat capacity and that all equals the energy (Q). but i don't know what to do then... there is something I am missing
 
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These are all about heat balance. Heat gained = heat lost (by different parts of the system). Heat = m*c*delta T.
 
delta T?
 
Change in temperature. Like in

boboclown said:
a cup of coffee (140g) cools from 75 degrees C down to 20. degrees C.

delta T = Tfinal - Tinitial = 75 - 20 = 55 deg C
 
oh ok ty
 
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