In order to have the ability to solve this problem, girlfriend will need to know the worth of water"s specific heat, i beg your pardon is detailed as

#c = 4.18"J"/("g" ""^

You are watching: A kilojoule is required to raise the temperature of how much water by 10.0 °c?

"C")#

Now, let"s assume that you don"t know the equation that permits you to plug in your values and also find just how much warmth would be essential to warmth that lot water through that numerous degrees Celsius.

Take a look in ~ the particular heat that water. Together you know, a substance"s specific heat tells you just how much heat is required in order to rise the temperature that #"1 g"# of that substance through #1^
"C"#.

In water"s case, you require to carry out #"4.18 J"# of warm per gram the water to boost its temperature by #1^
"C"# increase in a #"2-g"# sample the water. You"d need to administer it v

#overbrace("4.18 J")^(color(brown)("for 1 g the water")) + overbrace("4.18 J")^(color(brown)("for 1 g of water")) = overbrace(2 xx "4.18 J")^(color(green)("for 2 g that water"))#

To reason a #1^
"C"# rise in the temperature that #m# grams that water, you"d should supply it through

#overbrace("4.18 J")^(color(brown)("for 1 g that water")) + overbrace("4.18 J")^(color(brown)("for 1 g of water")) + " ,,, " = overbrace(m xx "4.18 J")^(color(green)("for m g that water"))#

This way that in order to rise the temperature of #m# grams that water by #n^
"C"#, you need to provide it with

#"heat" = m xx n xx "specific heat"#

This will certainly account for raising the temperature that the first gram that the sample by #n^
"C"#, and so on till you reach #m# grams the water.

And over there you have it. The equation that describes all this will for this reason be

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - warm absorbed#m# - the mass of the sample#c# - the certain heat the the substance#DeltaT# - the change in temperature, characterized as final temperature minus initial temperature

In your case, girlfriend will have actually

#q = 100.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^
"C")))) * (50.0 - 25.0)color(red)(cancel(color(black)(""^
"C")))#

#q = "10,450 J"#

Rounded to three sig figs and expressed in kilojoules, the answer will be

#"10,450" color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = color(green)("10.5 kJ")#