How much energy does it take to cool and heat up water

[pooker]

1. The problem statement, all variables and given/known data

1. How much energy does it take to heat 95.3 grams of water from -10 degrees celcius to 35 degrees celcius?
2. How much energy does it take to cool water from 125 degrees celcius to 75 degrees celcius?



2. Relevant equations
specific heat of water in solid is 2.06 j/g c
liquid is 6.01 j/g c
gas is 1.84 j/g c

delta h of fusion is 6.01 kj/mol
delta h of vap is 44 kj/mol





3. The attempt at a solution

1.

95.7(2.067)(10) = 1093.18 j = 10.93 kj
5.29(6.01) = 31.79 kj
95.3(4.184)(35) = 13955.77 j = 13.95 kj

answer is 56.7 kj

2
95.3(25)(1.84) = 4384 j = 43.84 kj
5.29(44) = 232.9 kj
25(95.3)(4.184) = 9968.38 j = 99.68 kj

answer is -376.42 kj

Now here is where I am confused. Is it -376.42 kj since it is an endothermic reaction? Or is only part of the reaction negative meaning my answer is completely different? Like when calculating delta h of evaporation would that be positive and the others negative? Or is it right that anytime you cool something off it is all completely negative? My book does not go into detail at all with this. It just gives one answer.

[Borek]

Think in terms of energy exchanged between your water and surroundings. To heat up the water it has to gain energy, hence its enthalpy goes up - and enthalpy change for water is positive. Quite the opposite for cooling.

[rowkem]

As Borek said - water has to gain or lose energy to heat and cool, respectively. The enthalpy change is always in relation to the system.

Heating: system gains energy: enthalpy change is positive
Cooling: system loses energy: enthalpy change is negative

Or is it right that anytime you cool something off it is all completely negative?

Yes, that is correct. Cooling an object is always negative, since it will be losing energy. In much the same way, heating an object is always positive.

If you're looking at successive heating and cooling, you would need to work out each cooling and heating segment separately, be weary of signs, and sum the enthalpies.