How much heat transfer to the skin by 25.0 g of steam onto the skin? The latent heat of vaporization of steam is L = 2.256*10^6 J/kg

Question

abb0t · Answer

Can't you use Q = mc$\Delta$T?

loser66 · Answer

yes, I have Q = mc delta t + mL 
I confused, not sure + or - mL

abb0t · Answer

condensation is exothermic. so +

loser66 · Answer

the steam releases the heat, so the heat-loss will have - sign, right?

abb0t · Answer

wait a minute. I thought i was in the chemistry section. this is physics. Oops.

abb0t · Answer

but since the substance is giving up energy in the form of heat, the enthalpy change of the substance is negative.

loser66 · Answer

let see :
skin temperature : 34
m = 25 *10^-3 kg
c =4190 J/kg.K
\(\triangle T= -66)
so, $$Q = 25*10^{-3}*4190*(-66)+25*10^{-3}*2.256*10^6 = 49486J$$

loser66 · Answer

but they said I am wrong

abb0t · Answer

Why is it -66?

loser66 · Answer

initial temperature is 100 , final is 34 so delta T = 34-100 =-66

abb0t · Answer

And did you keep your units consistent. Notice that you have kg and g's.

loser66 · Answer

25g = 25 *10^(-3) kg

abb0t · Answer

2256(25) J = 56400 J
25 g of 100°C hot water to water at 34 ° C gives off 
Q = cm$\Delta$T
c = 4.18 J per (g°K) 

Now can you finish it?

loser66 · Answer

Thank you, let me try. I got exactly this number , but it said "incorrect"
Ha!! just homework online. I can log out and ask my prof to get the reason.

abb0t · Answer

I don't have a calculator with me right now actually. I am doing this all on my flip phone. HAHA

raffle_snaffle · Answer

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raffle_snaffle · Answer

Q=(0.025kg)(2.256x1-^6)