How much heat is absorbed by pouring 10 grams of 100°C liquid water on a student’s hand. The water is cooled to body temperature, 37°C. Assume that all of the heat is absorbed by the hand
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
solve for q.
630J
To determine the amount of heat absorbed by the hand when pouring 10 grams of 100°C liquid water on it and cooling it to body temperature (37°C), we can use the specific heat capacity formula:
Q = mcΔT
Where:
Q is the heat absorbed or released
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 find the heat absorbed when cooling the water from 100°C to 37°C. The specific heat capacity of water is approximately 4.18 J/g°C:
Q1 = (10 g) * (4.18 J/g°C) * (37°C - 100°C)
Q1 = (10 g) * (4.18 J/g°C) * (-63°C)
Note that the change in temperature is negative because we are cooling the water.
Now, let's consider the heat absorbed by the hand to reach the temperature of the water (37°C). Assuming that all the heat from the water is absorbed by the hand, the hand's specific heat capacity can be approximated as the same value as water, 4.18 J/g°C:
Q2 = (10 g) * (4.18 J/g°C) * (37°C - 37°C)
Q2 = (10 g) * (4.18 J/g°C) * (0°C)
The change in temperature here is zero since the hand is at its initial temperature.
Now, we can calculate the total heat absorbed by the hand by adding Q1 and Q2:
Total heat absorbed = Q1 + Q2
Therefore, the heat absorbed by the hand when pouring 10 grams of 100°C water and cooling it to 37°C is Q1 + Q2. Please perform the calculations to get the specific value.