   Chapter 11, Problem 28P

Chapter
Section
Textbook Problem

How much thermal energy is required to boil 2.00 kg of water at 100.0°C into steam at 125°C? The latent heat of vaporization of water is 2.26 × 106 J/kg and the specific heat of steam is 2.010 J/(kg · °C).

To determine
Energy required to boil water at 100°C into steam at 125°C .

Explanation

Section 1:

To determine: Energy required to boil 100°C water into steam at 100°C .

Answer: Energy required to boil 100°C water into steam at 100°C is 4.52×106J .

Given Info:  Mass of water is 2.00kg , temperature of water is 100°C , and latent heat of vaporization is 2.26×106J/kg .

Formula to calculate the energy required to melt all the ice into liquid water is,

Qboil=mwaterLv

• Qboil is the energy required to boil the water 100°C water into steam at 100°C
• mwater is the mass of the water,
• Lv is the latent heat of vaporization,

Substitute 2.00 kg for mwater and 2.26×106J/kg for Lv to find Qboil .

Qboil=(2.00kg)(2.26×106J/kg)=4520000=4.52×106J

Therefore, the energy required to boil 100°C water into steam at 100°C is 4.52×106J .

Section 2:

To determine: Energy required to increase the temperature of steam from 100°C to 125°C .

Answer: Energy required to increase the temperature of steam from 100°C to 125°C

is 1.01×105J .

Explanation:

Given info: Mass of steam 2.00 kg, specific heat capacity of steam is 2010J/kg°C , Initial temperature of the steam is 100°C , and final temperature of steam is 125°C .

Formula to calculate energy required to increase the temperature of steam is,

Qsteam=msteamcsteam(Tf,steamTi,steam)

• Qsteam is the energy required to increase the temperature of steam,
• msteam is the mass of the steam,
• csteam is the specific heat capacity of steam,
• Ti,steam is the initial temperature of steam,
• Tf,steam is the final temperature of steam,

Substitute 2

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