We are given a thermodynamic process involving temperature changes. Let’s break down the problem step by step:
We start with the following temperature relationships:
We are given the temperature changes over a time interval and can use the following relationship between the initial and final temperatures:
\(T_0 = T\)
The first equation is based on the temperature change from \( 3T \) to \( 2T \) over 10 minutes:
\(\left(\frac{3T - 2T}{10}\right) = c_1 \left(\frac{3T + 2T}{2} - T\right) \) ....(\)
This equation is based on the fact that heat exchange rate is proportional to the temperature difference.
The second equation is based on the temperature change from \( 2T \) to \( T_f \) over 10 minutes:
\(\left(\frac{2T - T_f}{10}\right) = c_1 \left(\frac{2T + T_f}{2} - T\right) \) ....(i\)
Next, we need to take the ratio of the two equations (i) and (ii) to solve for the final temperature \( T_f \):
\(\frac{E(i)}{E(ii)} \Rightarrow \frac{\frac{T}{10}}{\frac{2T - T_f}{10}} = \frac{\frac{5T - 2T}{2}}{\frac{T_f}{2}}\)
After simplifying the equation, we get:
\(\frac{T}{2T - T_f} = \frac{3T}{T_f}\)
Rearranging this equation to solve for \( T_f \), we get:
\(T_f = 6T - 3T_f\)
Now, simplify further:
\(4T_f = 6T\)
Finally, we find the value of \( T_f \):
\(T_f = \frac{3}{2} T\)
The final temperature \( T_f \) is:
\(T_f = \frac{3}{2} T\)
A metal plate of area 10-2m2 rests on a layer of castor oil, 2 × 10-3m thick, whose viscosity coefficient is 1.55 Ns/m2. The approximate horizontal force required to move the plate with a uniform speed of 3 × 10-2ms-1 is:
In a Vernier caliper, \(N+1\) divisions of vernier scale coincide with \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is: