quantative energy

Energy Constants for Water (H₂O)

  • Heat Constants:

    • Heat of fusion (melting/freezing):

    • 334extJ/g334 ext{ J/g}

    • 79.7extcal/g79.7 ext{ cal/g}

    • Heat of vaporization (evaporating/condensing):

    • 2260extJ/g2260 ext{ J/g}

    • 540extcal/g540 ext{ cal/g}

    • Heat capacity (c) of liquid water:

    • 4.18extJ/g°C4.18 ext{ J/g°C}

    • 1.0extcal/g°C1.0 ext{ cal/g°C}

    • Heat capacity (c) of solid water:

    • 2.1extJ/g°C2.1 ext{ J/g°C}

    • 0.50extcal/g°C0.50 ext{ cal/g°C}

Problem Solving Strategies for Energy Calculations

  • For each thermal energy problem, sketch a warming or cooling curve to assist in deciding which equation(s) to use when solving the problem.

  • Maintain a reasonable number of significant figures in your answers.

Example Problems

Problem 1: Heating and Melting Ice

  • Question: How much energy must be absorbed by a 150 g sample of ice at 0.0°C that melts and then warms to 25.0°C?

  • Given Parameters:

    • Mass (M) = 150extg150 ext{ g}

    • Initial Temperature = 0.0°C0.0°C

    • Final Temperature = 25.0°C25.0°C

  • Calculation Steps:

    1. Calculate energy required for melting:

    • E<em>extmelt=MimesH</em>f=150extgimes334extJ/g=50100extJE<em>{ ext{melt}} = M imes H</em>f = 150 ext{ g} imes 334 ext{ J/g} = 50100 ext{ J}

    1. Calculate energy required for heating from 0.0°C to 25.0°C:

    • Eextheat=MimescimesriangleT=150extgimes4.18extJ/g°Cimes(25.0°C−0.0°C)E_{ ext{heat}} = M imes c imes riangle T = 150 ext{ g} imes 4.18 ext{ J/g°C} imes (25.0°C - 0.0°C)

    • =150imes4.18imes25.0=15675extJ= 150 imes 4.18 imes 25.0 = 15675 ext{ J}

    1. Total energy absorbed:

    • E<em>exttotal=E</em>extmelt+Eextheat=50100extJ+15675extJ=65775extJE<em>{ ext{total}} = E</em>{ ext{melt}} + E_{ ext{heat}} = 50100 ext{ J} + 15675 ext{ J} = 65775 ext{ J}

      • Final Answer:

    • Eexttotal=65775extJE_{ ext{total}} = 65775 ext{ J}

Problem 2: Energy Transfer in Icy Hot Lab

  • Question: A burner transfers 325 kJ of energy to 450 g of liquid water at 20°C. What mass of the water would be boiled away?

  • Given Parameters:

    • Energy transferred (E) = 325extkJ=325000extJ325 ext{ kJ} = 325000 ext{ J}

    • Mass of water (m) = 450extg450 ext{ g}

    • Final Temperature = 100°C100°C

  • Calculation Steps:

  • Use the heat of vaporization:

    • E=mimesHvE = m imes H_v

    • Rearranging gives: m=racEHvm = rac{E}{H_v}

    • Substitute the heat of vaporization value:

    • Hv=2260extJ/gH_v = 2260 ext{ J/g}

    • Calculate mass boiled away:

      • m=rac325000extJ2260extJ/g<br>ightarrowm=143.8extgm = rac{325000 ext{ J}}{2260 ext{ J/g}} <br>ightarrow m = 143.8 ext{ g}

    • Final Answer:

      • Mass of water boiled away = 143.8extg143.8 ext{ g}

Problem 3: Soft Drink Cooling

  • Question: How much energy needs to be removed from a 12 oz can of soft drink (assumed mass = 340 g) at 25°C to reach a freezer temperature of -12°C?

  • Given Parameters:

    • Mass (M) = 340extg340 ext{ g}

    • Initial Temperature = 25°C25°C

    • Final Temperature = −12°C-12°C

  • Calculation Steps:

    1. Change in Temperature (riangleTriangle T):

    • riangleT=T<em>extfinal−T</em>extinitial=−12°C−25°C=−37°Criangle T = T<em>{ ext{final}} - T</em>{ ext{initial}} = -12°C - 25°C = -37°C

    1. Calculate energy to remove:

    • E=MimescimesriangleT=340extgimes4.18extJ/g°Cimes(−37°C)E = M imes c imes riangle T = 340 ext{ g} imes 4.18 ext{ J/g°C} imes (-37°C)

    • =340imes4.18imes(−37)=−52638.68extJ= 340 imes 4.18 imes (-37) = -52638.68 ext{ J}

      • Final Answer:

    • Energy removed = −52638.68extJ-52638.68 ext{ J}, or approximately −53extkJ-53 ext{ kJ}.