In order to have the ability to solve this problem, you will require to recognize the value of water"s particular heat, which is noted as

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


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"C")#

Now, let"s assume that you don"t know the equation that enables you to plugin your values and find exactly how much warm would be required to warmth that much water through that plenty of degrees Celsius.

Take a look at the details heat of water. Together you know, a substance"s specific heat tells you just how much warm is necessary in order to boost the temperature the #"1 g"# of that substance by #1^
"C"#.

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

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

To cause a #1^
"C"# rise in the temperature the #m# grams of water, you"d must supply it with

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

This way that in stimulate to boost the temperature of #m# grams of water by #n^
"C"#, you need to provide it v

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

This will certainly account for enhancing the temperature that the first gram of the sample by #n^
"C"#, and also so on until you with #m# grams of water.

And there you have actually it. The equation that describes all this will thus be

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

#q# - warmth absorbed#m# - the mass of the sample#c# - the details heat the the substance#DeltaT# - the readjust in temperature, identified as final temperature minus initial temperature

In your case, you will have

#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")#


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