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…

Question

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature difference $(Δ T)$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80 ° C$ . The value of $t_{2}$ as mentioned in the graph will be

MARKS Content Team · Accepted Answer

Correct answer: 16. According to the Newton's law of cooling, -dT/dt=KT-T_0. Applying approximation, -(∆T)/(∆t)=K(T_1+T_2)/2-T_0 For first 6 min, T_1=80°C, T_2=40+20=60°C, T_0=20°C ⇒-(60-80)/6=K70-20⇒K=-(60-80)/(6×70-20) ⇒K=20/(6×50)=1/15 For 6 to t_2, T_1=60°C, T_2=20+20=40°C, T_0=20°C -(60-80)/(t_2-6)=K50-20⇒t_2-6=15×20/30⇒t_2=16 min

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature difference $(Δ T)$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80 ° C$ . The value of $t_{2}$ as mentioned in the graph will be

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Relevant Questions

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 80°C. The value of t2 as mentioned in the graph will be

Solution

Relevant Questions

In an experiment to verify Newton's law of cooling. a graph is plotted between. the temperature difference $(Δ T)$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80 ° C$ . The value of $t_{2}$ as mentioned in the graph will be