# Thermodynamics - ice->water->steam

## Homework Statement

A 8.0 g ice cube at -30 C is in a rigid, sealed container from which all the air has been evacuated.

How much heat is required to change this ice cube into steam at 240 C?

## Homework Equations

Q_net = 0 where Q is heat

C1 = Specific heat capacity, ice: 2.050 kJ/kg-K = 2050 J
C2 = Specific heat capacity, water: 4.184 kJ/kg-K = 4184 J
C3 = Specific heat capacity, water vapor: 1.996 kJ/-kgK = 1996 J

L_f = Latent heat fusion, water 334*10^3 J/kg
L_v = Latent heat vaporization, water 2257*10^3 J/kg

Q_net = Q1 + Q2 + Q3 = 0
Q1 = -30deg C ice --> 0deg C ice, Q2 = 0deg C water --> 100deg C water, Q3 = 100deg C vapor --> 240deg C vapor

Q_net = (C1)(M)(T1) + M(L_f) + (C2)(M)(T2) + M(L_v) + (C3)(M)(T3) = 0
M = 6.0 grams = 0.006 kg
T1 = 30 deg
T2 = 100 deg
T3 = 140 deg

## The Attempt at a Solution

Q_net = (C1)(M)(T1) + M(L_f) + (C2)(M)(T2) + M(L_v) + (C3)(M)(T3)

= (2050)(0.006)(30) + (0.006)(334*10^3) + (4184)(0.006)(100) + (0.006)(2257*10^3) + (1996)(0.006)(140)

= 369 + 2004 + 2510.4 + 13542 + 1995.16

= 20420.56 Joules

did i account correctly for all the phase changes etc? is my final answer correct? any help appreciated.

thanks

## Answers and Replies

Ygggdrasil
Science Advisor
Gold Member
It looks correct except m = 0.008kg, not 0.006kg.