How much heat is necessary to change 20g of ice at 0 degree C into water at 0 degree C? (Lf = 80kcal/kg)
is it 1.60kcal?
To calculate the amount of heat needed to change a substance from one state to another, you can use the formula:
Q = mass × latent heat
Where:
Q = heat energy (in calories or joules)
mass = mass of the substance (in grams or kilograms)
latent heat = latent heat of fusion (in calories or joules)
In this case, we are changing ice at 0 degrees Celsius into water at 0 degrees Celsius, so we are only considering the latent heat of fusion (Lf).
Let's calculate the heat:
Q = mass × Lf
Given:
mass = 20g
Lf = 80kcal/kg
First, convert 20g to kilograms:
mass = 20g ÷ 1000 = 0.02kg
Now, calculate the heat:
Q = 0.02kg × 80kcal/kg
Q = 1.6kcal
Therefore, the amount of heat necessary to change 20g of ice at 0 degrees Celsius into water at 0 degrees Celsius is 1.6kcal, which confirms that your initial answer is correct.
To determine the amount of heat required to change a given mass of ice into water at the same temperature, we need to use the equation:
Q = m × Lf
Where:
Q = Heat energy required (in calories)
m = Mass of substance being heated or cooled (in kg)
Lf = Latent heat of fusion (in calories per kg)
Given:
m = 20g = 0.02kg
Lf = 80kcal/kg
Converting kcal to calories, we have:
Lf = 80kcal/kg × 1000 cal/kcal = 80000 cal/kg
Now, we can plug the values into the equation:
Q = 0.02kg × 80000 cal/kg
Q = 1600 cal
Therefore, the amount of heat required to change 20g of ice at 0 degree C into water at 0 degree C is 1600 calories.
Yes. You have 0.02 kg. The answer is
(80 kcal/kg)(0.02 kg) = 1.6 kcal