How many joules of heat are given off when 5.00g of water cool from 348.0K to 298.0K?

Q =  -1045 J

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 348.0 K

Final temperature = 298.0 K

Heat given off = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g.K

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 298.0 K - 348.0 K

ΔT = - 50 K

Q = 5.0 g ×4.18 J/g.K× - 50 K

Q =  -1045 J

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :