Intro to Thermal Energy/Heat Transfer: Grade 11 Practice Problems

[-Dragoon-]

Does anyone have any good websites with practice problems and great in-depth explanations on thermal energy/heat transfer at the introductory level (grade 11)? I've searched many websites only to find questions that are too advanced. Questions like: What mass of steam is produced when 2.4 X 10^6 J of heat is applied to 4 kg of water at 20°C? Thanks in advance.

 

[rl.bhat]

What is the relevant equation for heat gained or lost by a body?

 

[-Dragoon-]

What is the relevant equation for heat gained or lost by a body?

Q = mcΔt, where Q is heat gained or lost, m is mass of the body, c is specific heat capacity and Δt is the change in temperature.

 

[rl.bhat]

Well.
The applied heat is utilized to increase the temperature of 4 kg of water from 20 degree Celsius 100 degree Celsius + m kg of water to convert into steam. I hope you know the latent heat of evaporation of water. Now proceed.

 

[-Dragoon-]

Well.
The applied heat is utilized to increase the temperature of 4 kg of water from 20 degree Celsius 100 degree Celsius + m kg of water to convert into steam. I hope you know the latent heat of evaporation of water. Now proceed.

The latent heat of vaporization for water is 2.3 X 10^6 (according to my book).

Q_total = M_waterC_waterΔt + M_waterLv_water
Q_total = (4 kg)(4200 J/(J·°C))(80°C) + (4 kg)(2.3 X 10^6 J / kg)
Q_total = 10544000 J. It takes 10544000 J to boil 4 kg of water that was initially at 20 degrees Celsius. Correct?


Thanks for all the help, I appreciate it.

 

[rl.bhat]

Qtotal = (4 kg)(4200 J/(J·°C))(80°C) + (4 kg)(2.3 X 10^6 J / kg)

The above step is not correct. Because all the water is not converted to steam.
Heat utilized to boil the water = 4*4200*80 = 1.344*10^6 Joule. Supplied heat is 2.4*10^6 Joule. So the excess heat is utilized to convert water to steam. Now find the mass of the water converted to steam.