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QUESTION

# To bring 1.0kg of water from 25 c to 99c takes how much heat input? (in kilojoules)

It will require "310,000 J", which is "310 kJ".

In order to answer this question, you you will need to use the following equation:

q = CmDeltaT,

where q is the quantity of heat gained or lost, in Joules or calories. I will use Joules (J). C is the capacity (of water in this case), m is mass in grams, and DeltaT is the difference in temperature, determined by subtracting the initial temperature, T_i from the final temperature, T_f.

Known/Given: C = "4.184 J/g·"^("o")"C" m = "1.0 kg" x "1000 g"/"1 kg" = "1000 g" T_i = "25"^("o")"C" T_f = "99"^("o")"C" DeltaT = "99"^("o")"C" - "25"^("o")"C" = "74"^("o")"C"

Unknown: q

Solution: q = CmDeltaT = "4.184 J/g·"^("o")"C" x "1000 g" x "74"^("o")"C" = "309,616 J" = "310,000 J"(due to the fact that 1.0 kg has two significant figures)

"310,000 J" x "1 kJ"/"1000J" = "310 kJ"