Can someone please help me with this first year chemistry question? (Enthalpy)? Please :)

A reaction raises the temperature of 40.0g of water by 5.00°C. What is the enthalpy change for the reaction?

The specific heat density of water are 4.18 J K-1 g-1 and 0.997g mL-1 respectively.

I know the answer is -836 J, i just don't know how to do it :( Can soomeone please explain to me how it is done? Answers greatly appreciated.

heat capacity is by def: C=change in enthalpy/change in Temperature

by rearranging we get: dH = C*dT the d's are supposed to del deltas

now we've got the specific heat given, which is the heat capacity pr mass. c = C/M

so C = 4.18 J K-1 g-1 * 40 g = 167.2 J/K

now using the the first def we get: dH = 167.2 J/K*5.00K (remember to use kelvin, but it doesn't matter here)

since this is a reaction that GIVES OFF heat, the answer will be negative.
You will get the answer.

cheers!

The reaction in question gave off heat, which we know because the water went up in temperature. That's the reason for the negative sign and the reaction is an exothermic one.

For the Joules:

q = (mass) (change in temp) (specific heat)

q = (40.0 g) (5.00°C) (4.18 J K-1 g-1) = 836 J

The °C and the K cancel because the °C is a _change_ in temperature, not a measurement of the temperature of an object. The "size" of one °C is equal to the "size" of one K, so they cancel.