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

To calculate the heat absorbed by a person's hand when 100 grams of liquid water at 100°C is poured on it and assuming the final temperature of the water is the normal body temperature of 37°C, we can use the equation:

Q = mcΔT

Where:
Q is the heat absorbed (in Joules)
m is the mass of the water (in kilograms)
c is the specific heat capacity of water (in Joules per gram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)

First, we need to convert the mass of the water from grams to kilograms:

Mass of water (m) = 100 grams = 0.1 kilograms

Next, we need to calculate the change in temperature:

ΔT = Final temperature - Initial temperature
ΔT = 37°C - 100°C
ΔT = -63°C

Now, we can calculate the heat absorbed:

Q = mcΔT
Q = 0.1 kg * c * -63°C

The specific heat capacity of water (c) is roughly 4.186 Joules per gram per degree Celsius. We can substitute this value into the equation:

Q = 0.1 kg * 4.186 J/g°C * -63°C

By multiplying the values together, we can find the heat absorbed by the person's hand.