A cylindrical copper rod of length 1.20 m and cross-sectional area 5.60 cm^2 is insulated to prevent heat loss through its surface. The ends are maintained at a temperature difference of 100 degrees C by having one end in a water-ice mixture and the other in boiling water and steam. Find the rate at which ice melts at one end (in grams/second).

To find the rate at which ice melts, we need to calculate the rate of heat conduction through the copper rod.

First, we can calculate the heat energy conducted through the rod using the formula:

Q = k * A * ΔT / L

Where:
Q = Heat energy conducted (in Joules)
k = Thermal conductivity of copper (approximately 400 W/mK)
A = Cross-sectional area of the rod (in square meters)
ΔT = Temperature difference between the ends of the rod (in Kelvin)
L = Length of the rod (in meters)

Let's calculate the values and substitute them into the formula:

k = 400 W/mK
A = 5.60 cm^2 = 5.60 * 10^-4 m^2
ΔT = 100 degrees Celsius = 100 Kelvin
L = 1.20 m

Q = (400 W/mK) * (5.60 * 10^-4 m^2) * (100 K) / (1.20 m)

Q = 0.1867 W

Now, we need to convert the heat energy into the amount of ice melted.

To do this, we need to know the heat of fusion of ice, which represents the amount of heat energy required to melt a certain mass of ice. The heat of fusion of ice is approximately 333.55 kJ/kg.

Let's convert the heat energy conducted to grams of ice melted:

Q = (mass of ice melted) * (heat of fusion of ice)

0.1867 W = (mass of ice melted) * (333.55 kJ/kg) / (1000 J/kJ)

mass of ice melted = (0.1867 W * 1000 J/kJ) / (333.55 kJ/kg)

mass of ice melted ≈ 0.559 grams per second

Therefore, the rate at which ice melts at one end of the rod is approximately 0.559 grams per second.

To find the rate at which ice melts at one end, we need to determine the rate of heat conducted through the cylinder and then use the heat of fusion of ice to calculate the mass melted per second.

First, let's find the thermal conductivity of copper:

The equation for heat transfer by conduction is given by:

Q = (k * A * ΔT * t) / L

Where:
Q = heat transferred
k = thermal conductivity
A = cross-sectional area
ΔT = temperature difference
t = time
L = length

Rearranging the equation, we get:

Q / t = (k * A * ΔT) / L

The rate at which heat is conducted through the cylinder is defined as:

H = Q / t

From given data:
Length of the copper rod (L) = 1.20 m
Cross-sectional area of the copper rod (A) = 5.60 cm^2 = 5.60 * 10^-4 m^2
Temperature difference (ΔT) = 100 °C
Time (t) = 1 second

Now, we need to find the thermal conductivity of copper (k). The thermal conductivity of copper is typically around 401 W/(m K).

Now, let's calculate the rate at which heat is conducted through the cylinder:

H = (k * A * ΔT) / L

H = (401 * 5.60 * 10^-4 * 100) / 1.20

H ≈ 186.33 W

Next, we need to calculate the rate at which ice melts. The heat of fusion of ice is the amount of heat required to convert a unit mass of ice at 0 °C into water at 0 °C. The heat of fusion of ice is commonly known as the latent heat of fusion of ice (L):

L = 334,000 J/kg

The rate at which ice melts (m) can be calculated using:

H = (m * L) / t

Rearranging the equation, we get:

m = (H * t) / L

Substituting the known values:

m = (186.33 * 1) / 334000

m ≈ 5.57 * 10^-4 kg

Since the density of ice is approximately 917 kg/m^3, we can calculate the rate at which ice melts in grams per second:

m = (5.57 * 10^-4) * (1000)

m ≈ 0.557 grams/second

Therefore, the rate at which ice melts at one end is approximately 0.557 grams/second.