Water has a specific heat of 4184 j/(kg k). how much energy is needed to heat a kilogram of water 4°c?

1 Answer
Junaid Mirza
Apr 13, 2018

#"16736 J"#

Explanation:

Use this equation

#"Q = mC"Δ"T"#

where

  • #"Q ="# Energy
  • #"m ="# Mass of sample
  • #"C ="# Specific heat of sample
  • #"ΔT ="# Change in temperature

#"Q" = 1 cancel"kg" × "4184 J"/(cancel"kg" cancel"K") × (4 -0) cancel"K" = "16736 J"#

#"Note: ΔT"_"Kelvin" = "ΔT"_"°C"#

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