How much energy will be required to heat a #"1.0 kg"# mass of water from #25^@"C"# to #99^@"C"#?

2 Answers
Feb 28, 2015

You'd need #"310 kJ"# to raise the temperature of 1 kg of water from #25# to #99^@"C"#.

So, you have all the information you need to use the equation

#q = m * c * DeltaT#, where

#q# - the amount of heat needed;
#m# - the mass of water - in your case 1.0 kg:
#c# - the specific heat of water;
#DeltaT# - the difference between the final temperature, #99^@"C"#, and the initial temperature, #25^@"C"#, of the water.

Plug your data into the equation to get

#q = "1,000 g" * 4.18"J"/("g" * ^@"C") * (99-25)^@"C"#

#q = "309,320 J" = "+309.3 kJ"#

Rounded to two sig figs, the answer will be

#q = "+310 kJ"#

Feb 28, 2015

Answer:

It will require #"310,000 J"#.

Explanation:

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, #c# is the specific heat capacity (of water in this case), #m# is mass in grams, and #DeltaT# is the difference in temperature, #DeltaT=T_"final"-T_"initial"#

Known/Given:
#c_"water"= 4.184 "J"/("g"*""^("o")"C"#

#m=1.0color(red)cancel(color(black)("kg"))xx"1000 g"/(1color(red)cancel(color(black)("kg")))=1.0xx10^3# #"g"#

#T_i="25"^@"C"#

#T_f="99"^@"C"#

#DeltaT="99"^@"C" - "25"^@"C"="74"^@"C"#

Unknown:
#q#

Solution:

Plug the known values into the equation and solve.

#q=4.184"J"/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx1.0xx10^3color(red)cancel(color(black)("g"))xx74^@color(red)cancel(color(black)("C")) = "310,000 J"# (rounded to two significant figures)