Under the spreading chestnut tree the village blacksmith dunks a red-hot horseshoe into a large bucket of 22*C water. How much heat was lost by the horseshoe in vaporizing .00100 kg. of water?
2.26 J
To determine the amount of heat lost by the horseshoe in vaporizing a certain amount of water, we need to use the equation
Q = mL,
where Q is the amount of heat lost or gained, m is the mass of the substance, and L is the latent heat of vaporization.
In this case, we are given the mass of water (m = 0.00100 kg) and the temperature change is from the initial temperature of the water (22°C) to its boiling point where it turns into steam (100°C).
First, let's calculate the temperature change:
ΔT = final temperature - initial temperature
= 100°C - 22°C
= 78°C
Next, we need to find the latent heat of vaporization of water (L). The latent heat of vaporization is the amount of heat required to convert a substance from a liquid to a gas at its boiling point, which is 100°C for water. The latent heat of vaporization for water is approximately 2,260 kJ/kg.
Now we can calculate the amount of heat lost (Q):
Q = mL
= 0.00100 kg × 2,260,000 J/kg
Converting kJ to J, we get:
Q = 2,260 J/g × 0.00100 kg
= 2.26 J
Therefore, the amount of heat lost by the horseshoe in vaporizing 0.00100 kg of water is 2.26 J.