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 18.5 g, and its initial temperature is -13.3 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.4 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand
Also we assume that all of the ice melts.
q1 = heat needed to raise T of cube at -13.3 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat needed to melt ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.
q3 = heat needed to raise T of melted ice from zero C to 28.4 C.
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Total q = q1 + q2 + q3
To calculate the amount of heat absorbed by the ice cube and resulting water, we can use the concept of heat transfer equation:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed or released (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
First, let's calculate the heat absorbed by the ice cube:
The mass of the ice cube is given as 18.5 g.
The initial temperature of the ice cube is -13.3 °C.
The final temperature is 0 °C since the ice cube melts into water.
Now, let's calculate the heat absorbed by the ice cube using the specific heat capacity of ice, which is approximately 2.09 J/g°C:
Q_ice = m_ice * c_ice * ΔT_ice
Q_ice = 18.5 g * 2.09 J/g°C * (0 °C - (-13.3 °C))
= 18.5 g * 2.09 J/g°C * 13.3 °C
= 517.9 J
Therefore, the ice cube absorbs 517.9 Joules of heat.
Next, let's calculate the heat absorbed by the resulting water:
The mass of the water is the same as the mass of the ice cube, which is 18.5 g.
The initial temperature of the water is 0 °C.
The final temperature of the water is 28.4 °C.
Now, let's calculate the heat absorbed by the water using the specific heat capacity of water, which is approximately 4.18 J/g°C:
Q_water = m_water * c_water * ΔT_water
Q_water = 18.5 g * 4.18 J/g°C * (28.4 °C - 0 °C)
= 18.5 g * 4.18 J/g°C * 28.4 °C
= 2192.858 J
Therefore, the resulting water absorbs 2192.858 Joules of heat.
To find the total heat absorbed by the ice cube and resulting water, we can simply add up the heat absorbed by each:
Total heat absorbed = Q_ice + Q_water
Total heat absorbed = 517.9 J + 2192.858 J
Total heat absorbed = 2710.758 J
Therefore, the ice cube and resulting water together absorb 2710.758 Joules of heat.