Question

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?

Answer 1

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 specific heat 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 specific heat 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 specific heat;
`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 sig figs, 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.