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"##