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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