An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80° to 70°C. The time taken to cool for the same object from 70°C to 60°C would be nearly :
- 10 min
- 12 min
- 20 min
- 15 min
The Correct Option is D
Solution and Explanation
\(\frac{80 - 70}{12} = K( \frac{80 + 70}{2} - 25) \,...(1)\)
\(\frac{70 - 60}{t} = K(\frac{70 + 60}{2} - 25) \,...(2)\)
on solving : \(t = 15\) min
Top Questions on Newtons Law of Cooling
The temperature of the body drops from 60°C to 40°C in 7 min. The surrounding temperature is 10°C. The temperature of the body drops from 40°C to T°C in 7 min. Find the value of T
- JEE Main - 2023
- Physics
- Newtons Law of Cooling
- In an experiment to verify Newton’s law of cooling, a graph is plotted between, the temperature difference (ΔT) of the water and surroundings and time as shown in figure. The initial temperature of water is taken as 80ºC. The value of \(t_2\) as mentioned in the graph will be ______.
- JEE Main - 2022
- Physics
- Newtons Law of Cooling
- In a room where the temperature is $30^{\circ}C$ a body cools from $61^{\circ}C$ to $59^{\circ}C$ in 4 minutes. The time taken by the body to cool from $51^{\circ}C$ to $49^{\circ}C$ will be about
- UPSEE - 2016
- Physics
- Newtons Law of Cooling
- Two spheres of different materials one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the larger sphere is $25$ minute and for smaller one is $16$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is
- BITSAT - 2016
- Physics
- Newtons Law of Cooling
- If the temperature of the body is increased by $ 10\%, $ the percentage increase in the emitted indication will be
- JEE Advanced - 2014
- Physics
- Newtons Law of Cooling
Questions Asked in NEET exam
- Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector \(\vec{P}\) of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :- NEET (UG) - 2024
- Magnetic Field
- Given below are two statements:
Statement I: Mitochondria and chloroplasts both double membranes bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast. I
In the light of the above statements, choose the mis appropriate answer from the options given below- NEET (UG) - 2024
- Cell structure and organelles
- Match List I with List II:Choose the correct answer from the options given below:
List I
List II
A Mesozoic Era I Lower invertebrates B Proterozoic Era II Fish & Amphibia C Cenozoic Era III Birds & Reptiles D Paleozoic Era IV Mammals - NEET (UG) - 2024
- Origin Of Life
- A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates :
- NEET (UG) - 2024
- Capacitors and Capacitance
- The terminal voltage of the battery, whose emf is\(10V\) and internal resistance \(1\), when connected through an external resistance of \(42\) as shown in the figure is:
- NEET (UG) - 2024
- Resistance and Resistivity
Concepts Used:
Newton’s Law of Cooling
Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings.
Derivation of Newton’s Law of Cooling
Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings.
If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,
dQ = ms dT2
The rate of loss of heat is given by,
dQ/dt = ms (dT2/dt) ……..(2)
Compare the equations (1) and (2) as,
– ms (dT2/dt) = k (T2 – T1)
Rearrange the above equation as:
dT2/(T2–T1) = – (k / ms) dt
dT2 /(T2 – T1) = – Kdt
where K = k/m s
Integrating the above expression as,
loge (T2 – T1) = – K t + c
or
T2 = T1 + C’ e–Kt
where C’ = ec