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The temperature of a body falls from 50oC to 40oC in 10 minutes. If the temperature of the surroundings is 20oC Then temperature of the body after another 10 minutes will be
- a)36.6oC
- b)33.3oC
- c)35oC
- d)30oC
Correct answer is option 'B'. Can you explain this answer?
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The temperature of a body falls from 50oC to 40oC in 10 minutes. If th...
Given information:
- Initial temperature of the body (T1) = 50°C
- Final temperature of the body (T2) = 40°C
- Time taken for the temperature to decrease (Δt) = 10 minutes
- Temperature of the surroundings (Tsur) = 20°C
Formula:
Newton's Law of Cooling states that the rate of change of temperature of a body is directly proportional to the difference between its temperature and the temperature of its surroundings. It can be represented as:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
where:
- dT/dt is the rate of change of temperature with respect to time
- T is the temperature of the body
- Tsur is the temperature of the surroundings
- k is a constant
Solution:
To find the temperature of the body after another 10 minutes, we need to first find the value of the constant 'k' using the given information.
Step 1: Find the temperature difference:
ΔT = T2 - T1
ΔT = 40°C - 50°C
ΔT = -10°C
Step 2: Substitute the values in Newton's Law of Cooling equation:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
\(\frac{dT}{dt} = -k(-10°C - 20°C)\)
\(\frac{dT}{dt} = 30k\)
Step 3: Substitute the values of Δt and ΔT in the equation:
\(\frac{ΔT}{Δt} = 30k\)
\(-\frac{10°C}{10 \text{ minutes}} = 30k\)
\(-1°C/\text{min} = 30k\)
Step 4: Solve for k:
k = \(\frac{-1°C/\text{min}}{30}\)
k = \(\frac{-1}{30}\) °C/min
Step 5: Substitute the value of k in the Newton's Law of Cooling equation:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
\(\frac{dT}{dt} = \frac{1}{30} (T_{sur} - T)\)
Step 6: Apply separation of variables and integrate the equation:
\(\int \frac{1}{T_{sur} - T} dT = \int \frac{1}{30} dt\)
ln|Tsur - T| = \(\frac{1}{30} t + C\)
Step 7: Apply initial condition to find the constant of integration:
When t = 0, T = T1
ln|Tsur - T1| = C
Step 8: Substitute the values back into the equation:
ln|Tsur - T| = \(\frac{1}{30} t + \text{ln}|Tsur - T1|\)
Step 9: Solve for T:
|Tsur - T| = e^(\(\frac{1}{30} t + \text{
- Initial temperature of the body (T1) = 50°C
- Final temperature of the body (T2) = 40°C
- Time taken for the temperature to decrease (Δt) = 10 minutes
- Temperature of the surroundings (Tsur) = 20°C
Formula:
Newton's Law of Cooling states that the rate of change of temperature of a body is directly proportional to the difference between its temperature and the temperature of its surroundings. It can be represented as:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
where:
- dT/dt is the rate of change of temperature with respect to time
- T is the temperature of the body
- Tsur is the temperature of the surroundings
- k is a constant
Solution:
To find the temperature of the body after another 10 minutes, we need to first find the value of the constant 'k' using the given information.
Step 1: Find the temperature difference:
ΔT = T2 - T1
ΔT = 40°C - 50°C
ΔT = -10°C
Step 2: Substitute the values in Newton's Law of Cooling equation:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
\(\frac{dT}{dt} = -k(-10°C - 20°C)\)
\(\frac{dT}{dt} = 30k\)
Step 3: Substitute the values of Δt and ΔT in the equation:
\(\frac{ΔT}{Δt} = 30k\)
\(-\frac{10°C}{10 \text{ minutes}} = 30k\)
\(-1°C/\text{min} = 30k\)
Step 4: Solve for k:
k = \(\frac{-1°C/\text{min}}{30}\)
k = \(\frac{-1}{30}\) °C/min
Step 5: Substitute the value of k in the Newton's Law of Cooling equation:
\(\frac{dT}{dt} = -k(T - T_{sur})\)
\(\frac{dT}{dt} = \frac{1}{30} (T_{sur} - T)\)
Step 6: Apply separation of variables and integrate the equation:
\(\int \frac{1}{T_{sur} - T} dT = \int \frac{1}{30} dt\)
ln|Tsur - T| = \(\frac{1}{30} t + C\)
Step 7: Apply initial condition to find the constant of integration:
When t = 0, T = T1
ln|Tsur - T1| = C
Step 8: Substitute the values back into the equation:
ln|Tsur - T| = \(\frac{1}{30} t + \text{ln}|Tsur - T1|\)
Step 9: Solve for T:
|Tsur - T| = e^(\(\frac{1}{30} t + \text{
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The temperature of a body falls from 50oC to 40oC in 10 minutes. If the temperature of the surroundings is 20oC Then temperature of the body after another 10 minutes will bea)36.6oCb)33.3oCc)35oCd)30oCCorrect answer is option 'B'. Can you explain this answer?
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