If the temperature of 1 gram of water changes from 22°C to 27°C, how many calories of heat were involved? How many joules?

1 Answer
Junaid Mirza
Jun 6, 2018

"5 cal, 20.92 J"5 cal, 20.92 J

Explanation:

Use this equation

"Q = mSΔT"Q = mSΔT

Where

  • "Q ="Q = Heat absorbed/released
  • "m ="m = Mass of sample
  • "S ="S = Specific heat of sample ("1 cal/g°C"1 cal/g°C or "4.184 J/g°C"4.184 J/g°C for water)
  • "ΔT ="ΔT = Change in Temperature

"Q" = 1 cancel"g" × 1 "cal"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "5 cal"

or

"Q" = 1 cancel"g" × 4.184 "J"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "20.92 J"

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