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
Jun 6, 2018

Answer:

#"5 cal, 20.92 J"#

Explanation:

Use this equation

#"Q = mSΔT"#

Where

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