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?
1 kJ = 0.24 kcal
c=4186 J/kg•K
Q=cmΔT
To calculate the amount of heat necessary to heat the water, we can use the specific heat capacity of water, which is approximately 1 kcal/kg°C (or 4.18 kJ/kg°C). The formula to calculate the heat energy is:
Q = m * c * ΔT
Where:
Q = heat energy (in calories or joules)
m = mass of the water (in kg)
c = specific heat capacity of water (in kcal/kg°C or kJ/kg°C)
ΔT = change in temperature (in °C)
Given:
m = 30 kg
c = 1 kcal/kg°C
First, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 80°C - 60°C
ΔT = 20°C
Now, substitute the given values into the formula:
Q = 30 kg * 1 kcal/kg°C * 20°C
Calculating the heat energy in kilocalories (kcal):
Q = 30 * 1 * 20 kcal
Q = 600 kcal
To convert this to kilojoules (kJ), we can use the conversion factor that 1 kcal = 4.18 kJ:
Q_kJ = 600 kcal * 4.18 kJ/kcal
Q_kJ = 2508 kJ
Therefore, it is necessary to provide 600 kcal (or 2508 kJ) of heat to raise the temperature of 30 kg of water from 60°C to 80°C.