Assuming the ambient temperature is 15 degrees Celsius, for a 310 ml can of water, and the final temperature of the water is 40 degrees Celsius, how many ml of water should be made boil ?
We usually call the ambient temperature the indoor air temperature, but considering only the air is equivalent to considering only the heat conduction between the water in the bottle and the air through the wall of the bottle. the bottle, but the propagation of heat It is not only like that. To put it simply, you turn on the air conditioner in a closed room and measure the ambient temperature at 26 degrees. But experience will tell you that you will always be cold at this temperature in winter. , but you won't be cold at this temperature in summer. Too fresh. The reason for this difference is that the wall temperature is low in winter and high in summer. Objects with the same temperature emit different amounts of heat per raythermally. difference.
High temperature objects It will radiate energy to low temperature objects. The water in the bottle will radiate energy to the wall at low temperature, so the temperature will decrease further, if heat is transmitted. Because the heat transfer is less than the heat lost by radiation, the internal energy of the water will become smaller and the temperature will decrease.
How much electricity is needed to cool 90 tons of water from 60 degrees to 15 degrees. degrees per day?
The heating power is 2000 watts, the raw water temperature is 15 degrees Celsius and after a time t, the temperature rises to T degrees Celsius.
Then, 2000*t=C*m*(T-15).
T=15+2000t/C*m.
Among them, C=4200J/(kg·℃), m is the mass of water, it depends on the situation specific .
Thus, the power is 2000 watts, the raw water temperature is 15 degrees Celsius, and m kg of eau, after the heating time t, the temperature rises to:
T=15 +t/(2.1*m)
The cooling of 90 tons of water 60 degrees to 15 degrees requires 4,726.35 kilowatt hours of electricity each day. A ton (1,000 kilograms) of water must release heat 4.2×10^3 Joules per kilogram of Celsius multiplied by 10^3 kilograms multiplied by one degree Celsius equals 4.2×10^6 Joules. 1 degree is 3.6 × 10^6 joules and 4.2 × 10^6 joules is 1.167 degrees. Heat transfer and work are equivalent in changing the internal energy of an object. 1 calorie of heat is equivalent to 4.2 joules of work.