What is the enthalpy change for the reaction? A reaction lowers the temp of 200g of water by 2.00C.?

Question

Specific Heat of water is 4.18 J K-1g-1.

a) 418 J
b) -418 kJ
c) 418 kJ
d) 1672 J
e) -1672 J

Answer 1

The answer is (e).

To calculate the enthalpy change, find the product of the specific heat of water, the mass of the water in grams and the temperature change of the water in kelvins. Since the change in degrees celcius equals the change in kelvins, use (-2).

E (enthalpy change) = 4.18*200*(-2) = -1678 J

I hope this helps!

Answer 2

ZachO is correct, as far as he went. What he calculated is the enthalpy change that the water experienced, and his result shows an exothermic process (the water lost heat).

But the question is, what was the enthalpy change for the reaction? If the water lost 1672 J of heat (DH=-1762 J), that heat went INTO the reaction. So, the reaction is an ENDOTHERMIC one with an enthalpy change of +1762 Joules.

Does that makes sense?