Quick question about the heat Q transferred between water and skin

AI Thread Summary
The discussion centers on the calculation of heat transfer (Q) between water and skin, specifically when water cools from 100°C to 34°C. The correct formula applied is Q = mcΔT, where the mass of water is 25.0 g and the specific heat capacity is 4190 J/kg°C. The user initially calculated Q as -6910 J, indicating heat lost by the water, while a source (Chegg) showed a positive value of 6910 J, reflecting heat gained by the skin. Clarification was provided that the question focuses on the heat gained by the skin, thus necessitating the use of the final temperature minus the initial temperature. The correct interpretation of the heat transfer direction is crucial for accurate results.
bluesteels
Messages
28
Reaction score
1
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
 
Physics news on Phys.org
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top