Thermodynamics: thermal energy?

Thermodynamics: thermal energy?

A small electric immersion heater is used to heat 100g of water for a cup of instant coffee. The heater is labeled “200 watts”, so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from 23 C to 100 C ignoring any thermal energy that transfers out of the cup.

The answer is 742 kJ. Can you show me the steps involved with solving this problem?

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1 Comment for Thermodynamics: thermal energy?

  • 1. kuiperbelt2003  |  February 15th, 2011 at 8:07 am

    the answer you quote is an energy, not a time…let’s follow this problem

    200W means the heater adds 200Joules of energy to the water each second

    the amount of energy needed to heat 100g of water from 23C to 100 C is given by:

    Q=mc(delta T) where m is the mass of the water, c is the specific heat, and delta T is the change of temp

    here, m=100 g
    c=4.2 J/gram/deg C
    delta T= 77

    thus, Q=100 g x 4.2 J/g/C x 77C = 32.3 KJ

    if the heater provides 200J/s, then the time needed to heat the water is

    t=32.3kJ/200J/s = 161.7 s

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