PHYSICS HEAT SERIES 4 (With Solutions)

WARI SECONDARY SCHOOL PHYSICS HEAT SERIES 4 - Solutions

Question 1

Estimate the rate at which ice melts in a wooden box 2cm thick of inside measurements 60 × 60 × 60cm. Assume that external temperature is 27°C and coefficient of thermal conductivity of wood = 0.1674 Js^-1m^-1°C^-1, latent heat fusion Ice = 336 × 10³ Jkg^-1

Solution:

We need to calculate the rate of heat transfer through the wood and then determine how much ice this heat will melt.

Heat transfer rate (Q/t) = k × A × ΔT / d

Where:
k = thermal conductivity = 0.1674 Js^-1m^-1°C^-1
A = surface area = 6 × (0.6m × 0.6m) = 2.16 m² (6 sides of cube)
ΔT = temperature difference = 27°C - 0°C = 27°C
d = thickness = 0.02 m

Q/t = (0.1674) × (2.16) × (27) / 0.02 = 488.2968 W

Now, calculate melting rate:

Q = mL ⇒ m/t = (Q/t)/L

Where L = latent heat of fusion = 336 × 10³ J/kg

m/t = 488.2968 / (336 × 10³) = 0.001453 kg/s = 1.453 g/s

Answer: 1.45 g/second

Question 2

A layer of Ice 10cm thick is formed on a pond. The temperature of air is -10°C. Calculate how long it will take for the thickness of ice to increase by 1mm. Density of ice = 1 g/cm³, thermal conductivity of ice = 0.005 cal s^-1cm^-1°C^-1, latent heat of ice = 80 cal/g.

Solution:

We need to calculate the time for the ice layer to grow by 1mm (0.1cm).

Heat conducted through ice: Q/t = k × A × ΔT / d

Where:
k = 0.005 cal s^-1cm^-1°C^-1
A = area (we can assume 1 cm² for simplicity)
ΔT = 0°C - (-10°C) = 10°C
d = ice thickness = 10 cm

Q/t = (0.005) × (1) × (10) / 10 = 0.005 cal/s

Mass of new ice formed per cm²: m = ρ × V = 1 g/cm³ × (1 cm² × 0.1 cm) = 0.1 g

Heat required to freeze this mass: Q = m × L = 0.1 g × 80 cal/g = 8 cal

Time required: t = Q / (Q/t) = 8 cal / 0.005 cal/s = 1600 s

Answer: 1600 seconds

Question 3

The temperature gradient in a 0.5m long rod is 80°C/metre. The temperature of the hot end of the rod is 30°C. Find the temperature of the cold end.

Solution:

Temperature gradient is the rate of change of temperature with distance.

Temperature gradient = ΔT / Δx = 80°C/m

Total temperature difference: ΔT = gradient × length = 80 × 0.5 = 40°C

Since the hot end is at 30°C, the cold end must be:

T_cold = T_hot - ΔT = 30°C - 40°C = -10°C

Answer: -10°C

Question 4

An electric heater is used in a room of total wall area 137m² to maintain a temperature of 20°C inside it when the outside temperature is -10°C. The walls have three (3) layers of different materials. The inner most layers is of wood of thickness 2.5cm, the middle layer is cement of thickness 1.0cm and the outer most layer is of brick of thickness 25.0cm. Find the power of electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125 Wm^-1°C^-1, 1.5 Wm^-1°C^-1 and 1.0 Wm^-1°C^-1 respectively.

Solution:

We need to calculate the total heat loss through the composite wall and determine the heater power needed to compensate.

Total thermal resistance R_total = R_wood + R_cement + R_brick

Where R = d / (k × A)

Calculate individual resistances per m²:

R_wood = 0.025 / 0.125 = 0.2 m²°C/W

R_cement = 0.01 / 1.5 ≈ 0.00667 m²°C/W

R_brick = 0.25 / 1.0 = 0.25 m²°C/W

Total resistance per m²: R_total = 0.2 + 0.00667 + 0.25 ≈ 0.45667 m²°C/W

Total temperature difference: ΔT = 20°C - (-10°C) = 30°C

Heat loss rate (power): P = A × ΔT / R_total = 137 × 30 / 0.45667 ≈ 9000 W

Answer: 9 kW

Question 5

A room is maintained at 20°C by a heater of resistance 20 ohms connected to 200V mains. The temperature is uniform throughout the room and heat is transmitted through a glass window of area 1m² and thickness 0.2cm. Calculate the temperature outside. Thermal conductivity of glass is 0.2 Cal s^-1 m^-1°C^-1 and mechanical equivalent of heat J=4.2J/cal.

Solution:

First calculate the power of the heater, then determine the outside temperature based on heat loss through the window.

Heater power: P = V²/R = (200)²/20 = 2000 W

Convert to calories/second: 2000 W = 2000 J/s = 2000/4.2 ≈ 476.19 cal/s

This power equals the heat loss through the window:

Q/t = k × A × ΔT / d

Rearrange to solve for ΔT:

ΔT = (Q/t) × d / (k × A) = 476.19 × 0.002 / (0.2 × 1) ≈ 4.7619°C

Outside temperature: T_out = T_in - ΔT = 20°C - 4.7619°C ≈ 15.24°C

Answer: 15.24°C

Dear students, with this exercise read seriously and intensively in principle of physics (Chand XI) from page 917-1085