Specific heat capacity calculation formula: C=Q/mΔT

C is the specific heat capacity: unit J/(kg-℃)

Q is the thermal energy: unit J

m is the mass: unit kg

t is the temperature: unit ℃

ΔT is the cumulative increase or decrease in temperature: unit ℃

Therefore, the work W required to raise the temperature ΔT is: Q*ΔT

W=Pt

W is work in Joules J

P is the power in watts W

t means time in seconds s

**Converting the work equation yields.**

The time required for heating t = W/P

The power required for heating P = W / t

Inquiry can be known that the specific heat capacity of water C = 4.2 * 10^3 J / kg – degrees Celsius

That is, the heat required to heat 1KG of water to rise 1 ℃ Q = C * m, that is, 4.2 * 10^3J

**Calculate how much power is needed to heat 1000L of water for half an hour to 100 degrees?**

Assume that the room temperature is 20 degrees, heating to 100 degrees, that is, 80 degrees of elevation

The density of water is 1, that is, 1000L of water for 1000Kg

0.5 hours is 1800s

Can be calculated: P = W / t = Q * ΔT / t = C * m * ΔT / t = 4.2 * 1000 * 1000Kg * 80 ℃ / 1800s = 16666W = 166KW

**The same can be calculated: 1000L of water using 30KW heating to 100 degrees need how much time?**

t=W/P=Q*ΔT/P=C*m*ΔT/P=4.2*1000*1000Kg*80℃/30000=11200s=3.11h

**The same can be calculated: 1000L water using 30KW heating to 60 degrees how much time it takes?**

t=W/P=Q*ΔT/P=C*m*ΔT/P=4.2*1000*1000Kg*40℃/30,000=5600s=1.56h

Extension: How many degrees of electricity is needed to heat 1000L of water from 20 degrees to 100 degrees

Knowledge: 1 degree electricity = 1KW*h=1000W*3600s=3.6*10^6J

Can be calculated: 4.2 * 1000 * 1000Kg * 80 ℃ / 3.6 * 10^6J = 93.3 degrees

The above is the theoretical value, the actual heat dissipation to be considered, the heating time needs more, the power needs to be greater