A body cools from 50.0°C to 48°C in 5s. How long will it take ...
Rate of cooling ∝ temperature difference between system and surrounding.
As the temperature difference is halved, so the rate of cooling will also be halved
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A body cools from 50.0°C to 48°C in 5s. How long will it take ...
°C to 20.0°C in 30 minutes. If the initial temperature of the surroundings is 20.0°C, what is the rate of heat loss from the body?
We can use the formula for rate of heat loss:
Q/t = kAΔT/L
where Q is the heat lost, t is the time, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and L is the thickness of the material.
Assuming the body is a uniform solid and the heat loss is mainly due to conduction, we can simplify the formula as:
Q/t = kA(θi - θf)/d
where θi is the initial temperature, θf is the final temperature, and d is the distance between the two surfaces.
We can rearrange the formula to solve for kA/d, which represents the heat transfer coefficient:
kA/d = Q/t/(θi - θf)
Substituting the given values, we get:
kA/d = (mCp)(θi - θf)/t/(θi - θs)
where m is the mass of the body, Cp is the specific heat capacity, and θs is the temperature of the surroundings.
Assuming the body is made of copper, which has a thermal conductivity of k = 401 W/mK, and has a density of ρ = 8960 kg/m³ and a specific heat capacity of Cp = 385 J/kgK, we can calculate the surface area and thickness of the body:
A = m/ρ = (1000 g)/(8960 kg/m³) = 0.1118 m²
V = m/ρ = (1000 g)/(8960 kg/m³) = 0.0001118 m³
d = V/A = 0.0001118 m³/0.1118 m² = 0.001 m = 1 mm
Substituting the values, we get:
kA/d = (1000 g)(385 J/kgK)(50.0 - 20.0)°C/(30 min)(50.0 - 20.0)°C/(20.0)°C
kA/d = 9.63 W/m²K
Therefore, the rate of heat loss from the body is:
Q/t = kAΔT/L = (9.63 W/m²K)(0.1118 m²)(50.0 - 20.0)°C/0.001 m
Q/t = 1.87 × 10³ W or 1.87 kW
Answer: The rate of heat loss from the body is 1.87 kW.
A body cools from 50.0°C to 48°C in 5s. How long will it take ...
Ans will be 5s not 10s. Apply newtons law of cooling formula 5s is the ans
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