SPECIFIC HEAT CALCULATION

Heat and Phase-Change Energy Calculations

2019-07-05 lixingdotcom Materials Science Fundamentals Leave a comment

Formula:

Calculation 1: Total Heat

A metal tank contains 20 kg of water at 20°C. The specific heat capacity of water is 4,186 J/(kg·°C). Ignoring heat exchange with the surroundings, how much heat is required to raise the water temperature to 50°C?

Solution:
Q=mc ΔT
Q=(2 kg)*(4186 joules/kilogram*digree celsius )*(50 digree celsius-20 digree celsius )
Q=251160 joules (heat required)

Calculation 2: Thermal Equilibrium

A metal tank contains 20 kg of water at 20°C. A 0.5 kg copper block at 90°C is placed in it. The specific heat capacities of water and copper are 4,186 and 387 J/(kg·°C), respectively. Ignoring heat exchange with the surroundings, what is the final equilibrium temperature?

Solution:
Assume an isolated system with no heat exchange with the surroundings.
Q=mc ΔT
Qc + Qw = 0 (conservation of energy)

(2 kg)*(4186 joules/kilogram*digree celsius )*(Tf-20 digree celsius )-( 0.5 kg)*(387 joules/kilogram*digree celsius )*(Tf-90 digree celsius ) =0
Tf = 21.58°C (final temperature)

Calculation 3: Phase Change, Example 1

A metal tank contains 20 kg of water at 20°C. Water has a specific heat capacity of 4,186 J/(kg·°C) and a latent heat of vaporization of 2,260,000 J/kg. Ignoring heat exchange with the surroundings, how much heat is required to turn the water into steam?

Solution:
Q=mc ΔT
Q=(2 kg)*(4186 joules/kilogram*digree celsius )*(100 digree celsius-20 digree celsius )
Q = 669760 joules (heat required)
This heat alone does not vaporize the water; it only heats it to 100°C. Additional heat is required for the phase change to steam.
We therefore need the additional heat associated with fusion (melting) or vaporization.
The phase-change equation is:
Q=mL
L is latent heat: energy absorbed or released during a phase change without a change in temperature.

The complete calculation is therefore:

Q = mcΔT (heat to reach 100°C) + mL (heat to vaporize the water)
Q=669760 joules + (2kg)*( 2260000 joules/kilogram )
Q = 5189760 joules (heat required to turn the water into steam)

Calculation 4: Phase Change, Example 2

A 3 kg block of ice is at -40°C. Specific heat capacities are 2,090 J/(kg·°C) for ice, 4,186 J/(kg·°C) for water, and 2,010 J/(kg·°C) for steam. The latent heats of vaporization and fusion are 2,260,000 J/kg and 333,000 J/kg, respectively. Ignoring heat exchange with the surroundings, how much heat is required to turn the ice into steam at a final temperature of 160°C?

Solution:
Q = heat to warm ice from -40°C to 0°C + heat of fusion + heat to warm water from 0°C to 100°C + heat of vaporization + heat to warm steam from 100°C to 160°C

Heat to warm ice from -40°C to 0°C
Q= (3kg)*( 2090 joules/kilogram )*(0 digree celsius – (-40 digree celsius ))=250800 joules
Heat of fusion (melting ice into water)
Q= (3kg)*( 333000 joules/kilogram ) =999000 joules
Heat to warm water from 0°C to 100°C
Q= (3kg)*( 4186 joules/kilogram )*(100 digree celsius – (0 digree celsius ))=1255800 joules
Heat of vaporization
Q= (3kg)*( 2260000 joules/kilogram ) =6280000 joules
Heat to warm steam from 100°C to 160°C
Q= (3kg)*( 2010 joules/kilogram )*(160 digree celsius – (100 digree celsius ))=361800 joules

Adding the terms:
250800 + 999000 + 1255800 + 6280000 + 361800 = 9147400 joules (heat required to turn the ice into steam at 160°C)