Question

1) If heating 1Kg of water, how much would the temperature increase by burning 100g of each substance each substance:

Answer 1

Answer:

a) The temperature of the water will rise by maximum 100 K, steam by 143.9 K

b) The temperature of the water will rise by maximum 100 K, steam by 1077.45 K

c) The temperature of the water will rise by maximum 100 K, steam by 972.75 K

d) The temperature of the water will rise by maximum 100 K, steam by 1027.86 K

e) The temperature of the water will rise by maximum 100 K, steam by 902.6 K

Explanation:

a) ΔH for ethanol = 29.65 kJ/g therefore, burning 100 g will produce;

29.65 × 100 = 2965 kJ

The specific heat capacity of water = 4.184 J/(g·K)

Therefore, 2965000= 1000 × 4.184 × ΔT

ΔT = 2965000 ÷ (1000 × 4.184) = 708.65 K

Latent heat of water = 2260 kJ/kg will be absorbed when the temperature reaches the boiling point of water hence we have

2965 - 2260 = 705 kJ to heat the water of which a maximum of 418.4 will boil the water and the steam temperature will rise by (705-418.4)/1.996 = 143.59 K

b) For Hexane: ΔH = 48.29 kJ/g

100 g will produce 4829 kJ

∴ Temperature change for the 1 kg water is given as follows

ΔT = 4829000 ÷ (1000 × 4.184) = 1154.16 K

However

4829 - 2260 = 2569

2569 - 418.4 = 2150.6

2150.6 / 1.996 = 1077.45 K

The final steam temperature will rise by 1077.45 K

c) For Kerosene(C₁₂H₂₆): ΔH = 46.2 kJ/g

100 g will produce 4620 kJ

∴ Hypothetically the temperature change for the 1 kg water is given as follows

ΔT = 4620000 ÷ (1000 × 4.184) = 1104.21 K

However

4620 - 2260 = 2360

2360 - 418.4 = 1941.6

1941.6 / 1.996 = 972.75 K

The final steam temperature will rise by 972.75 K

d) For Car Fuel(90% octane): ΔH = 47.3 kJ/g

100 g will produce 4730 kJ

∴ Temperature change for the 1 kg water is given as follows

ΔT = 4730000 ÷ 4184 = 1130.5 K

However

4730 - 2260 = 2470

2470 - 418.4 = 2051.6

2051.6 / 1.996 = 1027.86 K

The final steam temperature will rise by 1027.86 K

e) For Diesel (C₁₂H₂₃): ΔH = 44.8 kJ/g

100 g will produce 4480 kJ

∴ Temperature change for the 1 kg water is given as follows

ΔT = 4480000 ÷ 4184 = 1070.75 K.

However

4480 - 2260 = 2220

2220 - 418.4 = 1801.6

1801.6 / 1.996 = 902.6 K

The final steam temperature will rise by 902.6 K.