How much heat energy (in megajoules) is needed to convert 7 kilograms of ice at -9 degrees celsius to water at 0 degrees celsius? a. 3.09 mj b. 2.48mj c. 3.26 mj d. 2.13mj

B 2.48 mJ

Explanation:

The process can be broken into 2 steps

Step1: Heat required to raise the temperature of ice from -9 to 0 C

Q 1 = mcΔT

where m = mass of ice = 7 kg

c = specific heat capacity of ice = 2.108 kJ/kg-K

ΔT = T2-T1 = (273 - 264)K = 9 K

Step 2: Heat required for phase transition from ice at 0 C to water at 0C

Q 2= m*ΔH(fusion)

where ΔH(fusion) is the latent heat of fusion of ice = 334 kJ/Kg

Step 3: Total heat required

Q = Q1 + Q2 = 132.8+2338 = 2470.8 kJ ≅ 2.48 Mj

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 :

explanation:

0.07 g/cm³

1.37 g/cm³