# Specific & Latent Heat

## Homework Statement

How much heat is required to change 1.75 L of Ethyl Alcohol (C2H6O) at -50.0°C to a gas at its boiling point?

Ethyl Alcohol
V = 1.75 L
Ti = -50°C

46 g/mol
density = 0.789 g/cm3
boiling point = 78°C
specific heat (c)= 2400 J/kg*C°
Heat of Vaporization (Lv) = 850*103 J/kg

## Homework Equations

density m = ρV

specific heat Q = mcΔT

latent heat Q = mLv

## The Attempt at a Solution

First I determined the mass of the Ethyl Alcohol

m = ρV = (0.789 g/cm3) (1750 cm3) = 1380.75 g = 1.38075 kg

Then I solve for QNET

QNET = (mcΔT)l + mLv

QNET = [1.38 kg (2400 J/kg*C°) (78°C - (-50°C)] + [1.38 kg (850*103 J/kg)] = 1.60 * 106 J

According to the answer sheet, the answer is 4.24*105 J
What am I doing wrong?

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

## Answers and Replies

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
The answer sheet apparently has only accounted for the heat required to raise the temperature of the liquid ethanol from -50C to 78C.

So, did I do it correctly? It did say that "to a gas" . From my understanding of the problem as worded, I have to account for the amount of heat required to bring it to its boiling point, as well as the amount of heat required to turn it into a gas, right? Or did I do it incorrectly and I'm only suppose to calculate the amount of heat needed to bring it to the boiling point?

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
Only your professor knows for sure.