A 150KG STEEL ROD IN A BUILDING UNDER CONSTRUCTION SUPPORTS A LOAD OF 6050KG. DURING THE DSY THE RODS TEMP. INCREASES FROM 22C TO 47C CAUSING THE ROD TO THERMAALLY EXPAND AND RAISE THE LOAD 5.5mm. FIND THE EGERHU TRANSFER AS HEAT TO OR FROM THE ROD. THE SPECIFIC HEAT CAPACITY OF STEEL IS THE SAME AS FOR IRON.

b. FIND THE WORK DONE IN THIS PROCESS. IS WORK DONE ON OR BY THE ROD?
c. HOW GREAT IS THE CHANGE IN THE RODS INTERNAL ENERGY? DOES THE RODS INTERNAL ENERGY INCREASE OR DECREASE?

a. The energy transfer as heat to or from the rod is equal to the change in the rod's internal energy, which is equal to the product of the mass of the rod, the specific heat capacity of steel (which is the same as for iron), and the change in temperature. Therefore, the energy transfer is equal to 150 kg x 0.45 J/g°C x (47°C - 22°C) = 16,425 J.

b. The work done in this process is equal to the product of the load, the distance the load is raised, and the acceleration due to gravity. Therefore, the work done is equal to 6050 kg x 5.5 mm x 9.8 m/s2 = 3,093 J. The work is done by the rod.

c. The change in the rod's internal energy is equal to 16,425 J. The rod's internal energy increases.

To solve this problem, we need to apply the concepts of thermal expansion, heat transfer, work, and internal energy.

a. To find the energy transfer as heat to or from the rod, we can use the formula:

Q = mcΔT

Where:
Q is the energy transfer as heat
m is the mass of the rod
c is the specific heat capacity of steel/iron
ΔT is the change in temperature

First, let's find mc:

m = 150 kg (given)
c = specific heat capacity of steel/iron

The specific heat capacity of steel or iron is approximately 450 J/kg·°C.

mc = 150 kg * 450 J/kg·°C

Now, let's calculate ΔT:

ΔT = final temperature - initial temperature
ΔT = 47°C - 22°C

Substituting the values into the formula, we get:

Q = (150 kg * 450 J/kg·°C) * (47°C - 22°C)

b. To find the work done in this process, we need to calculate the displacement and use the formula:

Work = force * displacement

The force can be calculated by multiplying the load by the acceleration due to gravity:

Force = mass * acceleration due to gravity

Force = 6050 kg * 9.8 m/s^2

Now, let's calculate the displacement:

Displacement = 5.5 mm

Remember, we need the displacement in meters:

Displacement = 5.5 mm / 1000 (convert to meters)

Finally, calculate the work:

Work = Force * Displacement

c. The change in the rod's internal energy can be calculated using the formula:

ΔU = Q - W

Where:
ΔU is the change in internal energy
Q is the energy transfer as heat
W is the work done

Substitute the calculated values of Q and W into the formula to find the change in internal energy. The sign of ΔU will determine whether the internal energy increases or decreases. A positive value indicates an increase, while a negative value indicates a decrease.

Follow these steps and calculations to find the answers to the questions.

To solve this problem, we can use the formula for thermal expansion, as well as the formulas for heat transfer, work, and change in internal energy.

a. To find the heat transfer, we can use the formula:

Q = m * c * ΔT

Where:
Q = heat transfer
m = mass (150 kg)
c = specific heat capacity (same as for iron, which is 450 J/kg·°C)
ΔT = change in temperature (47°C - 22°C = 25°C)

Q = 150 * 450 * 25
Q = 1687500 J

Therefore, the heat transfer as heat to or from the rod is 1,687,500 Joules.

b. To find the work done, we can use the formula:

W = F * d

Where:
W = work done
F = force (load supported by the rod = 6050 kg * 9.8 m/s²)
d = displacement (5.5 mm, but we need to convert it to meters)

W = (6050 * 9.8) * (5.5 / 1000)
W = 33299 J

The work done is 33,299 Joules. Since the rod raised the load, work is done on the rod.

c. To calculate the change in the rod's internal energy, we can use the formula:

ΔU = Q - W

Where:
ΔU = change in internal energy
Q = heat transfer (1,687,500 J)
W = work done (33,299 J)

ΔU = 1,687,500 - 33,299
ΔU = 1,654,201 J

The change in the rod's internal energy is 1,654,201 Joules. Since the rod absorbed energy from its surroundings (heat transfer) and work was done on the rod, its internal energy increases.