how much heat is absorbed by pouring 10 grams of 100 degree C liquid water on a students hand. The water is cooled to body temperature 37 degree C
To determine the amount of heat absorbed by pouring 10 grams of 100 °C liquid water on a student's hand and cooling it to body temperature (37 °C), you need to calculate the heat transfer involved in the process.
The equation to calculate heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transfer (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
First, let's determine the heat transfer from the hot water to the student's hand. We know the mass of the water is 10 grams, and the initial temperature is 100 °C. However, we don't have the specific heat capacity of the hand. As an approximation, we can use the specific heat capacity of water, which is about 4.18 J/g°C.
Using the formula:
Q = m * c * ΔT
Q = 10 g * 4.18 J/g°C * (37 °C - 100 °C)
Q = 10 g * 4.18 J/g°C * (-63 °C)
Q = -2526 J (rounded to the nearest whole number)
The negative sign indicates that heat is being transferred from the water to the student's hand, resulting in a decrease in energy.
Therefore, approximately 2,526 Joules of heat will be absorbed by the student's hand during this process.