How much heat in joules is released when 45.0 g of water (C = 4.184 J/gºC) is cooled from 25.0°C to 23.0°C?

– 376.56 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Heat released (Q) =..?

Next, we shall determine the change in the temperature of the water. This can be obtained as follow:

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Change in temperature (ΔT) =.?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 23 – 25

Change in temperature (ΔT) = – 2 °C

Finally, we shall determine the heat released as follow:

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Change in temperature (ΔT) = – 2 °C

Heat released (Q) =..?

Q = MCΔT

Q = 45 × 4.184 × (– 2)

Q = – 376.56 J

Therefore, the heat released is – 376.56 J.

Note: The negative sign indicates that heat is released to the surrounding.

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 :