How much heat is required to raise 36 g ice at – 10.0oC to steam at 110oC? (get your answers from question #1) a) Amount of heat to go from –10.0oC to 0.00oC (solid, no phase change)

The total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

Explanation:

The heat involved in this process involves the following:

1. Heat to change ice at -10°C to ice at 0°C;

2. Heat to change ice at 0°C to water at 0°C

3. Heat to change water at 0°C to water at 100°C

4. Heat to change water at 100°C to steam at 100°C

5. Heat to change steam at 100°C to steam at 110°C

Specific heat capacity of ice, c = 2040 J/K/kg, Latent heat of fusion of ice, L = 3.35 × 10⁵ J/kg, specific heat capacity of water, c =  4182 J/K/kg, latent heat of vaporization of water, l = 2.26 × 10⁶ J/kg, specific heat capacity of steam, c = 1996 J/K/kg

Step 1: H = mcθ; where m = 30.0 g = 0.03 g, c = 2040 J/K/kg, θ = (0 -  -10) = 10 K

H = 0.03 * 2040 * 10 = 612 J

Step 2: H = mL, where  L = 3.35 × 10⁵ J/kg

H = 0.03 * 3.35 × 10⁵ = 10050 J

Step 3: H = mcθ, where c =  4182 J/K/kg, θ = (100 - 0) = 100 K

H = 0.03 * 4182 * 100 = 12546 J

Step 4: H = ml, where l = 2.26 × 10⁶

H = 0.03 * 2.26 × 10⁶ = 67800 J

Step 5: H = mcθ, where c = 1996 J/K/kg, θ = (110 - 100) = 10 K

H = 0.03 * 1996 * 10 = 598.8 J

Total heat required to raise ice at -10°C to steam at 110°C = (612 + 10050 + 12546 + 67800 + 598.8) J = 91606.8 J

Therefore, the total heat required to raise ice at -10°C to steam at 110°C = 91606.8 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 :

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