Calculate the heat produced from burning 10 kg of each type of fuel. Multiply the water converted to steam in kilograms (kg) by 2258kJ/kg, the heat of vaporization. The result of your calculation will be in kilojoules, mwater x 2258 kJ/kg = DQwater Wood - Fuel Mass (10kg), Water to Steam (90kg), Electricity (19KWh)
Coal - Fuel Mass(10kg), Water to Steam (140kg), Electricity (29KWh)
Oil - Fuel Mass (10kg), Water to Steam (190kg), Electricity (41KWh)
Methane - Fuel Mass (10kg), Water to Steam (240kg), Electricity (51KWh)
Please show the break down of each fuel.

Explanation:

1) For 10 kg of mass of wood

The mass of Water steam is 90kg

The heat produces by the fuel is

= 90kg x heat of vaporization

= ( 90kg x 2258kJ/kg)

= 203220kJ

2) For 10 kg of mass of coal

The mass of Water steam is 100kg

The heat produces by the fuel is

= 100kg x heat of vaporization

= ( 140kg x 2258kJ/kg)

= 316120kJ

3) For 10 kg of mass of wood

The mass of Water steam is 190kg

The heat produces by the fuel is

= 190kg x heat of vaporization

= ( 190kg x 2258kJ/kg)

= 429020kJ

4) For 10 kg of mass of wood

The mass of Water steam is 240kg

The heat produces by the fuel is

= 240kg x heat of vaporization

= ( 240kg x 2258kJ/kg)

= 541920kJ

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s