In a constant-pressure calorimetry experiment, a reaction gives off 22.8 kj of heat. the calorimeter contains 150 g of water, initially at 20.4°c. what is the final temperature of the water? the heat capacity of the calorimeter is negligibly small.

The final temperature of water is 56,7 ºC

Explanation:

We need to apply the specific heat capacity formula to solve this.

Q = m . C . ΔT

22,8 kJ =  150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

(We need to convert kJ in J to use the specific heat of water, be careful with units)

22,8 kJ = 22800 J (x1000)

22800 J  =  150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

22800 J = 627,9 J/ºC (Tfinal - 20,4º C)

22800 J / 627,9 ºC/ J = (Tfinal - 20,4º C)

36,3ºC = Tfinal - 20,4º C

36,3ºC + 20,4º C = 56,7 ºC

no but i do not usually comment or use my voice that much when it comes to social media

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.