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?

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

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

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

To calculate the final temperature of the water, you need to use the equation:

Q = mcΔT

Where:
Q is the heat absorbed by the water (343 kJ)
m is the mass of the water (43 kg)
c is the specific heat capacity of water (4.18 kJ/kg°C)
ΔT is the change in temperature (to be determined)

Rearranging the equation to solve for ΔT:

ΔT = Q / (mc)

Substituting the known values:

ΔT = 343 kJ / (43 kg * 4.18 kJ/kg°C)
ΔT ≈ 1.9°C

Therefore, the change in temperature is approximately 1.9°C.

To find the final temperature, you need to add this change to the initial temperature:

Final Temperature = Initial Temperature + ΔT
Final Temperature = 22.1°C + 1.9°C
Final Temperature ≈ 24.0°C

Hence, the final temperature of the water is approximately 24.0°C, not 1.9°C as you previously calculated.

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

deltatemp= 1.9K=1.9C

so, final temp= 22.1C+1.9C

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

This set up.

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