Physics word problem

Question

Indirect Cooling With Liquid Nitrogen. You are designing a system to cool an insulated silver plate of dimensions 2.00 cm × 2.00 cm x

0.45 cm. One end of a thermally insulated copper wire (diameter D = 3.00 mm and length L = 16.0 cm) is dipped into a vat of liquid

nitrogen (T = 77.2 K), and the other end is attached to the bottom of the silver plate.

1. If the silver plate starts at room temperature (67.0 °F), what is the initial rate of heat flow between the plate and the liquid nitrogen reservoir?
2. Assuming the rate of heat flow calculated in part (a), estimate the temperature of the silver plate after 30.0 seconds.

Answer 1

Solution: Indirect Cooling With Liquid Nitrogen

To solve this problem, we’ll break it down into two main parts:

1. Calculate the initial rate of heat flow between the plate and the liquid nitrogen reservoir using thermal conductivity.

2. Estimate the temperature of the silver plate after 30.0 seconds based on the rate of heat flow.

Part (a): Initial Rate of Heat Flow

Step 1: Given Information

1. Dimensions of Silver Plate:

  - Length (L_silver) = 2.00 cm = 0.02 m

  - Width (W_silver) = 2.00 cm = 0.02 m

  - Height (H_silver) = 0.45 cm = 0.0045 m

2. Dimensions of Copper Wire:

  - Diameter (D) = 3.00 mm = 0.003 m

  - Length (L_copper) = 16.0 cm = 0.16 m

3. Temperatures:

  - Initial temperature of the silver plate: T_initial = 67.0 °F = 19.4 °C = 292.6 K

  - Temperature of liquid nitrogen: T_LN2 = 77.2 K

4. Thermal Conductivities (typical values):

  - Copper: k_copper ≈ 390 W/m·K

Step 2: Calculate the Rate of Heat Flow (Fourier’s Law of Heat Conduction)

The rate of heat flow through the copper wire, Q/t, is given by:

Q/t = (k_copper * A * (T_initial - T_LN2)) / L_copper

1. Cross-sectional Area of Copper Wire (A):

  A = π * (D / 2)^2 = π * (0.003 / 2)^2 = 7.07 x 10^-6 m^2

2. Temperature Difference (ΔT):

  ΔT = T_initial - T_LN2 = 292.6 K - 77.2 K = 215.4 K

3. Substitute Values:

  Q/t = (390 * 7.07 x 10^-6 * 215.4) / 0.16 ≈ 3.71 W

Thus, the initial rate of heat flow between the silver plate and the liquid nitrogen reservoir is approximately 3.71 W.

Part (b): Estimate the Temperature of the Silver Plate After 30 Seconds

Step 1: Calculate Heat Capacity of Silver Plate

1. Volume of Silver Plate:

  V_silver = L_silver * W_silver * H_silver = 0.02 * 0.02 * 0.0045 = 1.8 x 10^-6 m^3

2. Mass of Silver Plate:

  - Density of silver (ρ_silver) ≈ 10,490 kg/m^3

  m_silver = ρ_silver * V_silver = 10,490 * 1.8 x 10^-6 = 0.0189 kg

3. Specific Heat of Silver (c_silver):

  c_silver ≈ 235 J/kg·K

4. Total Heat Capacity (C) of Silver Plate:

  C = m_silver * c_silver = 0.0189 * 235 = 4.44 J/K

Step 2: Temperature Change

The amount of heat transferred in 30 seconds is:

  Q = (Q/t) * t = 3.71 * 30 = 111.3 J

The temperature change (ΔT) is given by:

  ΔT = Q / C = 111.3 / 4.44 ≈ 25.1 K

Step 3: Final Temperature of Silver Plate

The estimated temperature of the silver plate after 30 seconds is:

  T_final = T_initial - ΔT = 292.6 K - 25.1 K = 267.5 K

Summary of Answers

- Initial rate of heat flow: ≈ 3.71 W

- Estimated temperature of the silver plate after 30 seconds: ≈ 267.5 K (or about -5.7 °C)