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Anonymous users2024-02-11I need your arguments! Gasoline or something! I don't remember.
It's in the book! That's how much energy a kilogram produces! Multiply the mass! It is equal to the energy consumed by lifting water m water * g * 10
If the gasoline internal combustion engine efficiency is at 30%.
The internal energy obtained from gasoline combustion is.
Due to the conversion efficiency! After being converted into mechanical energy, it is 4508000*30%=1352400j
The mass of the water raised is xkg
Then x* solution x= almost like this!!
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Anonymous users2024-02-10Ask them for help.
The internal energy obtained from gasoline combustion is.
q put = q suck.
The mass of the water raised is xkg
w=gh=mgh=x*x=
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Anonymous users2024-02-09First of all. The combustion value of gasoline is coke kg.
The efficiency of the general gasoline internal combustion engine is about 30%.
Words like this. The internal energy obtained from gasoline combustion is.
After being converted into mechanical energy in the internal combustion engine, it is 4508000*30%=1352400j, and the mass of water required to be 10m high is xkg
then x* solution yields x=13800
It seems that the end result is a little wrong.
That's the general idea.
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Anonymous users2024-02-08Summary. Hello, if 70% of the heat released by the complete combustion of a tank of gasoline is absorbed by water, how many kg of water temperature can be increased by 50oC??The heat released by the complete combustion of the fuel M1 is M1Q. <>
If 70% of the heat emitted by the complete combustion of a tank of gasoline is absorbed by water, how many kg of water temperature can be increased by 50 oC(Small.)
Hello, if 70% of the heat released by the complete combustion of a tank of gasoline is destroyed by the collapse of the car by water absorption, how many kg of water temperature can be raised by 50oc?The heat released by the complete combustion of the fuel M1 is M1Q. <>
According to the relevant data, all the heat released by the dissipation is absorbed by water m3, and the temperature t that water can rise is 10m27m3 t10m27m3 t <>
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Anonymous users2024-02-07Test Center: Calculation of calorific value.
Analysis & Process:
The calorific value of gasoline is the amount of heat emitted by the complete combustion of gasoline with a mass of 1kg.
The heat emitted when the gasoline is completely burned:
q=mq==
So: the heat emitted when the gasoline is completely burned is.
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Anonymous users2024-02-06The calorific value of gasoline is, and its physical meaning is: the heat emitted by the complete combustion of gasoline with a mass of 1kg is the heat released when gasoline is completely burned
q=mq==
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Anonymous users2024-02-05The heat released by the complete combustion of kerosene:
q=mq=2kg×
j/kg=j；q=cm t, the temperature at which water rises after absorbing heat: t=qc water. m water.
jj/(kg?℃)300kg
t=t0△t=14℃+73℃=87℃
A: The temperature of water can rise to 87
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Anonymous users2024-02-04The heat emitted by the complete combustion of gasoline:
q amp = mq = 10kg
j kg=j so the answer is:
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Anonymous users2024-02-03That must be the high calorie content of gasoline.
The calorific value of gasoline.
is j kg, dong zao hui or j g. Whereas, the heat capacity of water is j kg· , or 42 j g· . That's almost 1,100 times worse.
That is to say, the heat emitted by the complete combustion of 1 gram of gasoline can be used to heat up almost 1100 grams of water1, or about 11 grams of water from 0 to 100.
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Anonymous users2024-02-02The heat released by the complete combustion of diesel fuel q=mq=1kg, q=cmt, and the mass of water m=qc t=
200kg；
Therefore, C
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Anonymous users2024-02-0112000=20w
w=600kg
The calorific value of diesel fuel is 12000kcal kg
The specific heat of water is 1kcal kgk
It can raise the temperature of 600 kg of water to 20 degrees Celsius.
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