The calorific value of gasoline is 4.6 × 10 7 J / kg. When 210 g gasoline is burned completely, it can release () J heat. If all the heat is absorbed by water, it can make the gasoline more stable () kg of water, from 20 ℃ to 43 ℃, [whose specific heat capacity is 4.2 × 10 & # 179; J / (kg * ℃)]

The calorific value of gasoline is 4.6 × 10 7 J / kg. When 210 g gasoline is burned completely, it can release () J heat. If all the heat is absorbed by water, it can make the gasoline more stable () kg of water, from 20 ℃ to 43 ℃, [whose specific heat capacity is 4.2 × 10 & # 179; J / (kg * ℃)]

9.66x10 six times 100 4.6x10 seven times x0.21 = 9.66x10 six times 9.66x10 six times / (43-20) / 4.2x10 three times = 100=

The fuel is 93 gasoline, the calorific value is 4.5 times the 7th power joule of 10, the power of motorcycle is 6 kW, the price of 93 gasoline is 6.43 yuan per liter
The fuel is No.93 gasoline, the calorific value is 4.5 times the 7th power joule of 10, the motorcycle power is 6 kW, the price of No.93 gasoline per liter is 6.43 yuan, the quality of gasoline consumed by motorcycle and the fuel cost required are calculated

The energy consumed by a motorcycle for 60 minutes: e = 6 kW * 3600 seconds = 21600 joules. It takes 93 diesel oil mass (or volume) x = E / 4.5 E7, depending on the unit of calorific value. If x is volume, then the cost is 6.43 * x, if x is mass, the cost is 6.43 * x / fuel density

Nowadays, a kind of material called "solid alcohol" is commonly used as fuel in hot pot. It is known that the calorific value of this kind of fuel is 3 × 107 J / kg. If it is burned completely, how much J of heat can be released? If 21% of the heat released is absorbed by the water in the hot pot, how many kg of water can be heated by 80 ℃?

The results show that the heat released by the complete combustion of 0.2kg "solid alcohol" is: Q release = m alcohol, q = 0.2kg × 3 × 107j / kg = 6 × 106j; the heat absorbed by water is: Q absorption = 6 × 106j × 21% = 1.26 × 106j, ∵ Q absorption = cm △ T, ∵ mass of water: m water = q absorption, C △ t = 1.26 × 106j4.2 & nbsp; X 103J / (kg ·℃) × 80 ℃ = 3.75kg. A: the heat released by the complete combustion of 0.2kg "solid alcohol" is 6 × 106j. After 21% of the heat is absorbed by water, the water temperature rises by 80 ℃, and the mass of the water is 3.75kg

How much heat can be released by completely burning 21g of alcohol with calorific value of 3.0 times 10 to the 7th power J / kg? How many kilograms of water temperature can be raised by 50 ℃ by these heat

（21/1000）*3.0*10^7=630000J
630000/(4.2x10^3)/50=3kg