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Homework Help: Energy required to heat water

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A family uses 150 L of hot water each day. The hot water system is set to 75°C and the
    supply water has a temperature of 20°C. How much energy is required? If the hot water
    consumption was spread out evenly over the whole day (24 hrs), what is the total daily cost
    (assuming 14 c/kWh)?

    2. Relevant equations
    [itex]Q=mC \Delta T[/itex]

    [itex]C = 4.184 kJ/kg (HeatingCapacity)[/itex]
    3. The attempt at a solution
    Hi all, I have this question and I thought I was doing it correctly, but for some reason I'm not getting the required answer.
    [itex]density of water = 998 [/itex]

    [itex]Volume = 0.15 m^{3} [/itex]

    [itex]m = 149.7 kg [/itex]

    [itex]\Delta T = 55 [/itex]

    [itex] Q=(149.7)(4.184)(55) [/itex]
    [itex] Q=34424.26 kJ [/itex]

    [itex] seconds/day = 86400s [/itex]

    [itex] p = \frac{dQ}{dT} = \frac{\Delta Q}{\Delta T} [/itex]

    [itex] p = \frac{34424.26 kJ }{86400s}[/itex]

    [itex]p = 0.39843kW[/itex]

    [itex]p kWh = 0.39843kW (3600)[/itex]

    [itex]p kWh = 1434kWh[/itex]

    [itex]cost = (0.14c) (1434kWh)[/itex]

    [itex]cost = 200c = $2.00[/itex]

    However the answer states that it is $1.35/day

    I checked my work three times through, but I think I have a fundamental misunderstanding of something which is stopping me from getting the correct answer.
  2. jcsd
  3. May 26, 2014 #2


    User Avatar

    Staff: Mentor

    This is where you go wrong. In that second line, you are actually calculating how much energy is used every hour:
    0.39843\ \textrm{kW} \times \frac{3600\ \textrm{s}}{\textrm h} = 1434\ \textrm{kJ/h}
    Instead, find the conversion factor to go from kJ to kWh:
    1 \textrm{kWh} &= 1 \textrm{kW} \times \textrm{h} \\
    &= 1 \frac{\textrm{kJ}}{ \textrm{s}} \times \textrm{h} \\
    &= 1 \frac{\textrm{kJ}}{ \textrm{s}} \times \textrm{h} \times \frac{3600\ \textrm{s}}{\textrm h} \\
    &= 3600\ \textrm{kJ}
  4. May 26, 2014 #3
    Ahh, I plugged that conversion factor in after I found Q, then got $1.33... close enough?

    Cheers for the response
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