A 43 kg sample of water absorbs 343 kJ of heat. If the water was initially at 22.1°C, what is its final temperature?

Question

I got 1.9 but i think i didn't do it correctly

Answer 1

343kj=43kg*4.18kj/kg(deltaTemp)

deltatemp= 1.9K=1.9C

so, final temp= 22.1C+1.9C

Answer 2

thanks.

what about:

block of aluminum with a volume of 98.2 cm3 absorbs 67.4 J of heat. If its initial temperature was 32.5 °C, what is its final temperature? (density of aluminum = 2.70 g/cm3)

I know i have to change the volume to grams. but i don't know how to lay out the problem

Answer 3

Look at the set up I did. There's no layout at all. Substitute the numbers and solve.

Answer 4

huh? what set up that u did? wait r u bobpursley?

Answer 5

This set up.
http://www.jiskha.com/display.cgi?id=1284585197
mass Al = volume x density for that part.