Question

the rate of change of temperature of an object is proportional to the difference between the temperature of the object and its environment a glass of hot water at a temperature of 70 degree celsius is kept in a room temperature which is at a temperature of 30 degree Celsius if after three minutes the temperature of the water is 50 degree Celsius what will be its temperature after 5 minutes​

Answer 1

The temperature of object after 5 min = 41.66°C

Step by step explanation:

It is given that the rate of temperature of an object is directly proportional to the difference between temp. of object and environment.

It means

 $\frac{\partial T}{\partial t} \propto (T_1 - T_2)$                                              ( i )

Removing proportionality sign by a constant K

$\frac{\partial T}{\partial t} = K (T_1 - T_2)$                                            ( ii )

Integrating equation(ii)

$\int\limits \partial T= T = \int\limits^a_b K (T_1 - T_2) \partial t$                           (iii)

Putting T = 50° C

Initial temperature of object = T₁= 70° C                 (change with time)

Room temperature T₂  = 30° C                                 (constant)

a=3 min

b = 0 min   in  equation (iii)

50 = K (70 - 30) {3 - 0} = 120 K

So, K =    $\frac{\text{\large 50}}{\text{\large 120}} = \frac{\text{\large 5}}{\text{\large 12}}$      

After finding the value of K

Now T₁ becomes 50° C

and Room temperature T₂  = 30° C

a= 0 min

and b = 5 min

So equation (iii) becomes

$\textbf{\Large T = } \frac{\textbf{\Large 5}}{\textbf{\Large 12 }} \textbf{\Large(50-30)(5-0)}$

$\textbf{\Large T = } \frac{\textbf{\Large 5}}{\textbf{\Large 12 }} \textbf{\Large(20)(5)}\\\\\textbf{\Large T = } \frac{\textbf{\Large 500}}{\textbf{\Large 12 }}$

$\textbf{\Large The temperature of object after 5 min =}$

$\textbf{\Large T = } \frac{\textbf{\Large 125}}{\textbf{\Large 3 } }\textbf{\Large = 41.66}$ ° C

Answer 2

Given :

rate of change of temperature is directly proportional to difference between the temperature of object and its environment.

initial temperarture = 70 degree celsius.

temperature after 3 minutes = 50 degree celsius.

ambient temperature = 30 degree celsius.

To find:

temperature after 5 minutes.

Solution :

$\frac{dT}{dt}$  ∝ $T- T_{a}$

$\frac{dT}{dt}$ = k(T - 30) , where k is constant of proportionality.

integrate the above equation;

we get ,      $ln$(T-30) = kt +c ,

where c is constant of integration.

 substitute  T = 70 at t = 0;

c = $ln (40)$

2) at t = 3mins T = 50

$ln(50-30) = k(3) + ln40$

$ln20-ln40 =3k\\ln\frac{20}{40} = 3k\\ln\frac{1}{2} = 3k\\k = \frac{1}{3}ln\frac{1}{2}$

we need to find T at t = 5 mins

put t =5 in equation

$ln(T-30)= \frac{1}{3}ln\frac{1}{2} + ln40$

T = 25.89 degree celsius.