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## Homework Statement

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

## Homework Equations

density m = ρV

specific heat Q = mcΔT

latent heat Q = mL

_{v}

## 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!