What is the change in temperature for #"30 mL"# of water that was heated with #"2450 J"# of thermal energy? Assume the specific heat capacity stays constant within this temperature range.

1 Answer

#"19.50 °C"#

Explanation:

Density of water = #("1g") /("mL")#

Specific heat of water = # ("4.186 J") / ("g °C")#

#"Mass of water = Density of water × Volume of water"#

#= "1 g"/(cancel"mL") × 30 cancel"ml" = "30 g"#

#"Q" = "mS"Delta"T"#

#Delta"T" = "Q" / ("mS")#

#Delta"T" = ("2450 J") / ("30 g" × ("4.186 J") / ("g °C"))#

#Delta"T = 19.50 °C"#