How much heat is absorbed by a persons hand if 100 grams of liquid water at 100°C is poured on his hand? (Assume the final temperature of the water will be normal body temperature of 37°C)
What idiot would pour 100C water on his hands?
heat=masswater*specifiheatcapacitywater*(67)
To calculate the amount of heat absorbed by a person's hand when liquid water at 100°C is poured on it, we need to use the heat transfer equation:
Q = m * c * ΔT
Where:
Q is the heat absorbed (in joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.184 J/g°C), and
ΔT is the change in temperature (in °C).
First, we need to calculate the change in temperature (ΔT). The initial temperature of the water is 100°C, and the final temperature will be 37°C (normal body temperature). Therefore, ΔT is:
ΔT = 37°C - 100°C = -63°C
Next, we plug the values into the equation:
Q = 100g * 4.184 J/g°C * -63°C
Calculating this equation will give us the amount of heat absorbed by the person's hand when the water is poured on it.