What is the final temperature of water if 50.5 grams was water, initially at 35 degrees Celsius, that absorbed 800 J of energy?

1 Answer
Junaid Mirza
May 29, 2018

#"38.78°C"#

Explanation:

Use this equation

#"Q = mSΔT"#

Where

  • #"Q ="# Amount of heat added/liberated
  • #"m ="# Mass of sample
  • #"S ="# Specific heat of sample (#"4.186 J/g°C"# for water)
  • #"ΔT ="# Change in temperature of sample

Heat is added to water. So, final temperature (#"T"#) is greater than initial one.

#"ΔT" = "Q"/"mS"#

#"(T – 35°C)" = "800 J"/"50.5 g × 4.186 J/g°C"#

#"(T – 35°C) = 3.78°C"#

#"T = 3.78°C + 35°C = 38.78°C"#

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