How much heat is absorbed when the temperature of water is 50 ℃ and the mass is 100 g? The specific heat capacity of water is 4.2 × 10 & sup3; (10 cubic) coke / (kg ·℃) ② Algebra ③ Get the solution

How much heat is absorbed when the temperature of water is 50 ℃ and the mass is 100 g? The specific heat capacity of water is 4.2 × 10 & sup3; (10 cubic) coke / (kg ·℃) ② Algebra ③ Get the solution

Q = cm * Δ t = 4.2 × 10 & sup3; coke / (kg ·℃) * 0.1kg * 20 ·℃ = 8400 J

Under 1 standard atmospheric pressure, the mass of water is 400g, and the temperature is 55 ℃. After absorbing 84000j of heat, how much ℃ does the temperature increase?

Q＝cm(t-t0)
　　84000J=4200J/(kg•℃) x 0.4kg x ( t-55℃)
　　t-55℃ =50℃
　　t=105℃
Because the boiling point of water is 100 ℃ at 1 standard atmospheric pressure, the water temperature remains unchanged after boiling,
So at last, the water temperature can only rise to 100 ℃, and the temperature only rises by 100 ℃ - 55 ℃ = 45 ℃
A: the temperature rises by 45 degrees Celsius

How much does the temperature rise when a kilogram of 20 ℃ water absorbs the quintic angle heat of 2.1 times 10

Q=mc△t
210000=1*4200*△t
△t=210000/4200=50°C
Temperature rise 50 ° C
Now it's 50 + 20 = 70 ° C

When the mass of water is 2kg and the temperature is 15 degrees, after absorbing the heat of 1.68 * 1000000j, how much does the temperature rise to?

Q=cm(t2-t1)
t2-t1=Q/(cm)=1.68*100000/[(4.2*1000)*2)]=20
The temperature rises to T2 = T1 + 20 = 15 + 20 = 35 degrees

How much heat does a ton of water need to increase by 1 degree?
What is the heat of 1 ton of 0.5MPa (about 150 ℃) pressure steam?

1 ton of water to increase 1 degree needs heat = 1000 * 4200 * 1 = 4.2 * 10 ^ 6J = 1000 kcal
The heat of 1 ton 0.5MPa (about 150 ℃) pressure steam mainly includes latent heat of vaporization and sensible heat
The heat you asked should be enthalpy, = 1000 * 1kg, corresponding to the steam enthalpy value under the temperature and pressure. For the specific value, please refer to the steam enthalpy and entropy table. I don't remember

When a ton of water rises from 20 ℃ to 60 ℃, how much heat does it need for every 1 ℃ rise in temperature, and how much natural gas does it total? I use it to calculate energy saving!

The heat calculation formula: q = cm △ t = 4.2 × 103J / (kg ℃) × 1000kg × (60-20) ℃ = J
1 liter = 1 cubic decimeter, and the mass m of the water is 1kg, which is 4200j

One ton of water needs less heat to rise by one degree

Q=cmt
=4.2X10^3J /(kg·℃)*1000kg*1℃
=4.2X10^6J

Heat requirement for 1 cubic meter of water to increase 1 degree

2 × 10 & sup 3; kilojoules per cubic meter of water

How much heat does 1kg of water need per liter

Q=CM△t
=4.2X10^3J/(kg·℃)x1kgx1℃
= 4.2x 10^3J

How many calories does a gram of water increase by 1 degree

It is known that the weight of water is one gram (0.001 kg), the specific heat capacity of water is 4200 coke / kg. Degree, and the temperature is 1 degree
According to the formula, heat = mass x specific heat capacity x temperature, then:
The heat absorption capacity of water = 0.001x4200x1 = 4.2j
A: it takes 4.2 joules of heat to raise a gram of water by 1 degree