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# How much heat, in joules and in calories, must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius?

You would need to remove ##"-1300 Joules"##, or ##"320 cal"##, worth of heat from that much water to get its temperature to drop by ##"10"^@"C"##.

First, start by listing the of water in both calories per gram Celsius, and in Joules per gram Celsius.

##c_("water") = "1.00""cal"/("g" * ^@"C") = "4.186""J"/("g" * ^@"C")##

A substance's will tell you how much energy you need to put in to increase the temperature of 1 gram of that substance by 1 degree Celsius.

Now, the relationship between heat added or removed and the change in temperature that consequently takes place is described by this equation

##q = m * c * DeltaT##, where

##q## - the amount of heat removed; ##m## - the mass of the substance - in your case, water; ##c## - water's ; ##DeltaT## - the change in temperature.

Since the equation uses grams, use water's molar mass to go from moles to grams

##"1.75"cancel("moles water") * "18.0 g"/("1"cancel("mole water")) = "31.5 g water"##

Now just plug your values into the equation and solve for ##q##

##q_("calories") = "31.5"cancel("g") * "1.00""cal"/(cancel("g") * ^@cancel("C")) * (15 - 25)^@cancel("C") = "-315 cal"##

##q_("Joules") = "31.5"cancel("g") * "4.186""J"/(cancel("g") * ^@cancel("C")) * (15 - 25)^@cancel("C") = "-1316.7 J"##

Rounded to two , the number of sig figs given for 25 and 15 degrees Celsius, the answers will be

##q_("calories") = color(red)("-320 cal")##

##q_("Joules") = color(red)("-1300 J")##

**SIDE NOTE** The answers have a negative sign because heat is being lost by the system.