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20 September, 23:06

What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid water is 4.18 J/gºC.

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  1. 20 September, 23:26
    0
    amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC with specific heat of liquid water is 4.18 J/gºC

    = 4.18 J/gºC * 80g * (75-45) ºC

    = 10032J
  2. 20 September, 23:30
    0
    =4.18*80 * (75-45)

    =4.18*80*30

    =10032
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