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 -11.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.0 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.

Heat absorbed by ice raising T from -11.4 to zero.

q1 = mass ice x specific heat ice x (Tf-Ti).
q1 = 16.0g x 2.087 J/g x 11.4 = ?J

q2 = heat to melt the ice to liquid at zero C.
q2 = mass x heat fusion ice.
q2 = 16.0g x 333.6 J/g = ?

q3 = heat to raise T of melted ice at zero C to 31.0 C.
q3 = mass H2O x specific heat H2O x (Tf-Ti)
q3 = 16.0g x 4.184 J/g*C x 31 =?

Total heat absorbed = q1 + q2 + q3 = ?

Do you have the heat of fusion? The specific heat of ice? The specific heat of H2O?

Enthalpy of fusion

333.6 J/g
6010. J/mol

Specific heat of solid H2O (ice)
2.087 J/(g·°C)
37.60 J/(mol·°C)

Specific heat of liquid H2O (water)
4.184 J/(g·°C)
75.37 J/(mol·°C)

Specific heat of gaseous H2O (steam)
2.000 J/(g·°C)
36.03 J/(mol·°C)

To calculate the amount of heat absorbed by the ice cube and resulting water, we can use the equation:

Q = m * c * ΔT

where:
Q is the heat absorbed (in Joules)
m is the mass of the ice cube (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (in °C)

First, let's calculate the change in temperature of the ice cube. Since the ice melts and reaches the temperature of your skin (31.0 °C), the change in temperature for the ice cube is:

ΔT1 = 31.0 °C - (-11.4 °C)
= 42.4 °C

Next, we calculate the heat absorbed by the ice cube using the equation:

Q1 = m * c * ΔT1

Q1 = 16.0 g * 4.18 J/g°C * 42.4 °C

Q1 ≈ 2843.52 J

Now that the ice cube has melted and the resulting water has reached the temperature of your skin, we can calculate the heat absorbed by the resulting water. Since the water reaches 31.0 °C and its initial temperature was -11.4 °C, the change in temperature for the water is:

ΔT2 = 31.0 °C - (-11.4 °C)
= 42.4 °C

Using the same equation as before, we calculate the heat absorbed by the water:

Q2 = m * c * ΔT2

Q2 = 16.0 g * 4.18 J/g°C * 42.4 °C

Q2 ≈ 2843.52 J

Therefore, the total heat absorbed by the ice cube and resulting water is the sum of Q1 and Q2:

Q total = Q1 + Q2
≈ 2843.52 J + 2843.52 J
≈ 5687 J

So, approximately 5687 Joules of heat is absorbed by the ice cube and resulting water.