(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

How much heat is required to change 1.75 L of Ethyl Alcohol (C_{2}H_{6}O) at -50.0°C to a gas at its boiling point?

Ethyl Alcohol

V = 1.75 L

T_{i}= -50°C

46 g/mol

density = 0.789 g/cm^{3}

boiling point = 78°C

specific heat (c)= 2400 J/kg*C°

Heat of Vaporization (L_{v}) = 850*10^{3}J/kg

2. Relevant equations

density m = ρV

specific heat Q = mcΔT

latent heat Q = mL_{v}

3. The attempt at a solution

First I determined the mass of the Ethyl Alcohol

m = ρV = (0.789 g/cm^{3}) (1750 cm^{3}) = 1380.75 g = 1.38075 kg

Then I solve for Q_{NET}

Q_{NET}= (mcΔT)_{l}+ mL_{v}

Q_{NET}= [1.38 kg (2400 J/kg*C°) (78°C - (-50°C)] + [1.38 kg (850*10^{3}J/kg)] = 1.60 * 10^{6}J

According to the answer sheet, the answer is 4.24*10^{5}J

What am I doing wrong?

Thank you in advance! Any and all help is appreciated!

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# Homework Help: Specific & Latent Heat

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