You wish to heat 30 kg of water from 60°C to 80°C. How many kcal of heat are necessary to do this? To how many kJ does this correspond?
To calculate the amount of heat required to raise the temperature of an object, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy in joules (J),
m is the mass of the object in kilograms (kg),
c is the specific heat capacity of the substance (water in this case) in joules per kilogram per degree Celsius (J/kg°C),
ΔT is the change in temperature in degrees Celsius (°C).
Given:
Mass of water, m = 30 kg
Initial temperature, T1 = 60°C
Final temperature, T2 = 80°C
First, let's calculate the change in temperature:
ΔT = T2 - T1
ΔT = 80°C - 60°C
ΔT = 20°C
The specific heat capacity of water is approximately 4.18 J/g°C or 4.18 kJ/kg°C.
Since the mass of water is given in kilograms, we can use this value directly.
Now, substituting the given values in the formula:
Q = mcΔT
Q = 30 kg * 4.18 kJ/kg°C * 20°C
Q = 2,508 kJ
Thus, the number of kilocalories (kcal) of heat required is 2,508 kcal, and this corresponds to 2,508 kilojoules (kJ).