A bucket full of water is kept in a room and it cool from 75° C t
A. T<sub>1</sub> = T<sub>2</sub> = T<sub>3</sub>
B. T<sub>1</sub> < T<sub>2</sub> < T<sub>3</sub>
C. T<sub>1</sub> > T<sub>2</sub> > T<sub>3</sub>
D. T<sub>1</sub> < T<sub>3</sub> < T<sub>2</sub>
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
CONCEPT:
- When there is a temperature difference between two points then heat flow occurs from higher temperatures to lower temperatures. This phenomenon is called heat transfer.
- Newton’s law of cooling: It states that the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings .i.e., the rate of cooling ∝ ΔT
\(\Rightarrow \frac{{dT}}{{dt}} = -k\left( {\Delta T} \right)\)
Where dT/dt = rate of heat transfer or heat loss per sec (J/s), (ΔT = T - To) = temperature difference, and k is some constant
EXPLANATION:
- According to Newton’s law of cooling
\(\Rightarrow \frac{{dT}}{{dt}} = -k\left( {T-T_o} \right)\)
\(\Rightarrow \frac{{dT}}{{\left( {T-T_o} \right)}} = kdt\)
On integrating left sides from T2 to T1 and right side from t to 0, we get
\(\Rightarrow ln(\frac{T_2-T_o}{T_1-T_o})=-kt\)
As we know, room temperature is around 25°C, therefore To = 25°C
CASE 1: When T2 = 75°C and T1 = 70° C
- Time taken to cool done is
\(\Rightarrow t_1=-\frac{1}{k}ln(\frac{T_2-T_o}{T_1-T_o})=-\frac{1}{k}ln(\frac{75-25}{70-25})\)
\(\Rightarrow t_1=-\frac{1}{k}ln(\frac{50}{45})=-ln(1.1111)=0.105 \,sec\)
CASE 2: When T2 = 70°C and T1 = 65° C
- Time taken to cool done is
\(\Rightarrow t_2=-\frac{1}{k}ln(\frac{T_2-T_o}{T_1-T_o})=-\frac{1}{k}ln(\frac{70-25}{65-25})\)
\(\Rightarrow t_1=-\frac{1}{k}ln(\frac{45}{40})=-ln(1.125)=0.118 \,sec\)
CASE 3: When T2 = 65°C and T1 = 60° C
- Time taken to cool done is
\(\Rightarrow t_2=-\frac{1}{k}ln(\frac{T_2-T_o}{T_1-T_o})=-\frac{1}{k}ln(\frac{65-25}{60-25})\)
\(\Rightarrow t_1=-\frac{1}{k}ln(\frac{40}{35})=-ln(1.428)=0.133 \,sec\)
From above it is clear that t1 < t2 < t3.