Quick question about the heat Q transferred between water and skin

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SUMMARY

The discussion centers around the calculation of heat transfer (Q) between water and skin, specifically using the formula Q=mcΔT. The correct calculation for heat transferred by 25.0 g of water cooling from 100°C to 34°C is Q=0.025 * 4190 * (34-100) = -6910 J, indicating heat lost by water. The confusion arises from interpreting the direction of heat transfer, where the heat gained by the skin is considered positive, thus the value should be reported as 6910 J.

PREREQUISITES
  • Understanding of the specific heat capacity (c) of water, which is 4190 J/(kg·°C).
  • Familiarity with the concept of heat transfer and the first law of thermodynamics.
  • Knowledge of the formula Q=mcΔT for calculating heat transfer.
  • Basic understanding of temperature scales and conversions.
NEXT STEPS
  • Review the principles of thermodynamics, focusing on heat transfer mechanisms.
  • Study the specific heat capacities of different substances for comparative analysis.
  • Learn about the implications of negative and positive values in heat transfer calculations.
  • Explore practical applications of heat transfer in real-world scenarios, such as in cooking or engineering.
USEFUL FOR

Students studying thermodynamics, physics educators, and professionals in fields related to heat transfer and thermal dynamics.

bluesteels
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Homework Statement
How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue to assume that the skin temperature does not change.

Assume that water and steam, initially at 100∘C, are cooled down to skin temperature, 34∘C, when they come in contact with your skin
Relevant Equations
Q=mc delta
c = 4190
My work

Q= 0.025 * 4190 * (34-100) = -6910

but on chegg but did they do

Q= 0.025 * 4190 * (100-34) = 6910. I thought the initial is 100C and final is 34 because it goes from 100 to 34
 
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bluesteels said:
Homework Statement:: How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue to assume that the skin temperature does not change.

Assume that water and steam, initially at 100∘C, are cooled down to skin temperature, 34∘C, when they come in contact with your skin
Relevant Equations:: Q=mc delta
c = 4190

My work

Q= 0.025 * 4190 * (34-100) = -6910

but on chegg but did they do

Q= 0.025 * 4190 * (100-34) = 6910. I thought the initial is 100C and final is 34 because it goes from 100 to 34
You did the correct calculation, final minus initial, for the heat gained by the water, which is negative. That is not what the question asks for.
 
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