A 49kg sample of water absorbs 336kJ of heat.If the water was initially at 25.1Celsius, what is its final temperature?

Question

jfraser · Answer

same as your last question, just solve for deltaT, not Q

anonymous · Answer

can you show me how i would perform the question? sorry.

anonymous · Answer

$$336KG\frac{ ? }{ ? }$$should it be this? >$$\DeltaT=    336KG$$

anonymous · Answer

SORRY IT WONT LET ME BACK SPACE.

anonymous · Answer

should the answer be 0.273192943, when rounded 0.27 or 0.3

anonymous · Answer

49kg x 336kJ x 25.1 C=?

whpalmer4 · Answer

$$Q=mc\Delta{t}$$$$\Delta{t} = \frac{Q}{mc}$$
$$\Delta{t} = \frac{336kJ}{49kg*c}$$
(I don't recall the value of c, you'll have to supply it)
Remember to add $\Delta{t}$ to the original temperature to get the final temperature.

whpalmer4 · Answer

By the way, look at the units on your proposed multiplication.  Does mass*energy*temperature have the right dimensions?