Question

how much heat is needed to change 10g of ice at -10 degree celcius to 10g steam at 110 degree celcius?

Answer 1

Answer:

these are the steps... Q is heat fyi..

1) heat the ice to it's melting point of 0C... Q = (m cp dT) ice

2) melt the ice at 0 C.. Q = m x ∆Hfus

3) heat the resulting water to the boiling point  of 100C..Q=(m Cp dT) liquid

4) vaporize the liquid at 100C.. Q = m x ∆Hvap

5) heat the vapor from 100 to 110 C.... Q = (m Cp dT) vapor..

got all that?.. let Q' = total heat...

Q' = (m cp dT) ice + m x ∆Hfus + (m Cp dT) liquid + m x ∆Hvap + (m Cp dT) vapor

plug and chug..

Q' = 10gx(0.5 cal/gC)x(15C) + 10gx(80 cal/g) + 10gx(1cal/gC)x(100C) + 10gx(540cal/g) + 10gx(0.5 cal/gC)x(10C)

Q' = 75 cal + 800 cal + 1000 cal + 5400 cal + 50 cal

Q' = 7325 cal = 7.325 kcal

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one comment.. notice that I switched unit from cal/gK to cal/gC?

I can do this because the heat capacity is based on a temperature change.  ∆T.. and ∆T in K = ∆T in C... right? Because C + 273 = K.. so it would go something like this...

let two temps be 1 and 2.. C1 is temp 1 in degrees C.. same for C2.. anyway

C1 - C2 = (C1 + 273) - (C2 + 273) = K1 - K2...

...Afficher plus

Explanation: