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 16.0 g, and its initial temperature is -10.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.3 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.
q1 = heat absorbed by ice from -10.4 to zero.
q1 = mass ice x specific heat solid ice x (Tfinal-Tinitial).
Note: Tfinal = 0 and Tinitial = -10.4
q2 = heat absorbed to melt ice at zero to liquid at zero.
q2 = mass ice x heat fusion ice
q3 = heat absorbed in water at zero moving to skin temperature.
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial).
Total heat absorbed = q1 + q2 + q3.
To determine the amount of heat absorbed by the ice cube and resulting water, we can use the equation:
Q = mcΔT
Where:
Q is the amount of heat absorbed,
m is the mass of the substance (in this case, either the ice cube or resulting water),
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
First, let's determine the heat absorbed by the ice cube:
Since the ice cube melts and then reaches the temperature of your skin, we need to consider two separate processes: the heat absorbed during the melting phase and the heat absorbed during the subsequent temperature increase.
1. Heat absorbed during melting phase:
To calculate the heat absorbed during the melting phase, we can use the equation:
Q(melting) = m × ΔHf
Where:
Q(melting) is the heat absorbed during the melting phase,
m is the mass of the ice cube, and
ΔHf is the heat of fusion, which is the amount of heat required to change a substance from the solid to the liquid phase.
The heat of fusion for water is 334 J/g.
So, Q(melting) = 16.0 g × 334 J/g = 5,344 J
2. Heat absorbed during temperature increase:
Now let's calculate the heat absorbed during the temperature increase. To do this, we use the equation mentioned earlier:
Q = mcΔT
In this case, we need to consider the mass and initial temperature of the resulting water, as well as the change in temperature.
The specific heat capacity of water is 4.18 J/g°C.
So, Q(temperature increase) = (16.0 g + mass of resulting water) × 4.18 J/g°C × (28.3°C - (-10.4°C))
Simplifying the equation, we have:
Q(temperature increase) = (16.0 g + mass of resulting water) × 4.18 J/g°C × 38.7°C
Now, to find the mass of the resulting water, we need to equate the heat absorbed during the melting phase to the heat absorbed during the temperature increase:
5,344 J = (16.0 g + mass of resulting water) × 4.18 J/g°C × 38.7°C
Solving for mass of resulting water, we get:
mass of resulting water = (5,344 J) / (4.18 J/g°C × 38.7°C) - 16.0 g
Finally, substitute the value of the mass of resulting water back into the equation for Q(temperature increase) to find the heat absorbed during the temperature increase.
So, the total heat absorbed, Q, is the sum of Q(melting) and Q(temperature increase).