To treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 12.6 g, and its initial temperature is -10.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.1 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand. Constants may be found here.
To determine the amount of heat absorbed by the ice cube and the resulting water, we need to use the formula:
Q = mcΔT
Where:
Q is the heat absorbed or released
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's determine the heat absorbed by the ice cube:
The initial temperature of the ice cube is -10.4 °C, and the final temperature is 0 °C (since ice melts at 0 °C). Therefore, the change in temperature (ΔT) of the ice cube is:
ΔT = 0 °C - (-10.4 °C) = 10.4 °C
The specific heat capacity of ice is 2.09 J/g°C. So, we can calculate the heat absorbed by the ice cube using the formula:
Q_ice = m_ice * c_ice * ΔT
Substituting the given values:
Q_ice = 12.6 g * 2.09 J/g°C * 10.4 °C
Q_ice ≈ 273 J
Now, let's determine the heat absorbed by the resulting water:
The initial temperature of the water is 0 °C, and the final temperature is 28.1 °C. Therefore, the change in temperature (ΔT) of the water is:
ΔT = 28.1 °C - 0 °C = 28.1 °C
The specific heat capacity of water is 4.18 J/g°C. So, we can calculate the heat absorbed by the resulting water using the formula:
Q_water = m_water * c_water * ΔT
Since all the water remains in your hand, the mass of the water is equal to the mass of the ice cube, which is 12.6 g. Substituting the given values:
Q_water = 12.6 g * 4.18 J/g°C * 28.1 °C
Q_water ≈ 1500 J
Therefore, the total heat absorbed by the ice cube and resulting water is approximately:
Q_total = Q_ice + Q_water
Q_total = 273 J + 1500 J
Q_total ≈ 1773 J