Question

The temperature of equal masses of three different liquids x, y and z are 10\(^{\circ}\)C, 20\(^{\circ}\)C and 30\(^{\circ}\)C respectively. The temperature of mixture when x is mixed with y is 16\(^{\circ}\)C and that when y is mixed with z is 26\(^{\circ}\)C. The temperature of mixture when x and z are mixed will be :

• A. 20.28\(^{\circ}\)C
• B. 23.84\(^{\circ}\)C
• C. 25.62\(^{\circ}\)C
• D. 28.32\(^{\circ}\)C

Hint:

In calorimetry problems involving multiple mixing scenarios, the goal is often to find the ratios of specific heats first. Once you have the ratios, you can solve for the final unknown temperature. Always set up the "Heat Lost = Heat Gained" equation carefully.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a calorimetry problem. We need to find the final temperature of a mixture of liquids x and z, given the results of mixing x with y, and y with z. The masses of the liquids are equal.
Step 2: Key Formula or Approach:
The principle of calorimetry states that when two substances at different temperatures are mixed, the heat lost by the hotter substance is equal to the heat gained by the colder substance, assuming no heat loss to the surroundings. The formula for heat transfer is \(Q = ms\Delta T\), where `m` is mass, `s` is specific heat capacity, and \(\Delta T\) is the change in temperature.
Step 3: Detailed Explanation:
Let the equal mass of each liquid be `m`, and their specific heats be \(s_x, s_y, s_z\).
Initial temperatures: \(T_x = 10^\circ C\), \(T_y = 20^\circ C\), \(T_z = 30^\circ C\).
Case 1: Mixing x and y
Final temperature is 16\(^{\circ}\)C. Liquid y loses heat, and liquid x gains heat.
Heat lost by y = Heat gained by x
\[ m s_y (T_y - 16) = m s_x (16 - T_x) \] \[ s_y (20 - 16) = s_x (16 - 10) \] \[ 4s_y = 6s_x \implies 2s_y = 3s_x \implies s_y = \frac{3}{2}s_x \] Case 2: Mixing y and z
Final temperature is 26\(^{\circ}\)C. Liquid z loses heat, and liquid y gains heat.
Heat lost by z = Heat gained by y
\[ m s_z (T_z - 26) = m s_y (26 - T_y) \] \[ s_z (30 - 26) = s_y (26 - 20) \] \[ 4s_z = 6s_y \implies 2s_z = 3s_y \] Case 3: Mixing x and z
Let the final temperature be \(T_f\). Liquid z will lose heat, and liquid x will gain heat.
Heat lost by z = Heat gained by x
\[ m s_z (T_z - T_f) = m s_x (T_f - T_x) \] \[ s_z (30 - T_f) = s_x (T_f - 10) \] To solve for \(T_f\), we need a relationship between \(s_x\) and \(s_z\). From the first two cases:
We have \(2s_z = 3s_y\) and \(s_y = \frac{3}{2}s_x\). Substitute \(s_y\) into the equation for \(s_z\):
\[ 2s_z = 3\left(\frac{3}{2}s_x\right) = \frac{9}{2}s_x \implies 4s_z = 9s_x \implies s_z = \frac{9}{4}s_x \] Now substitute this into the equation for the third case:
\[ \left(\frac{9}{4}s_x\right) (30 - T_f) = s_x (T_f - 10) \] Cancel \(s_x\) from both sides:
\[ \frac{9}{4} (30 - T_f) = (T_f - 10) \] \[ 9(30 - T_f) = 4(T_f - 10) \] \[ 270 - 9T_f = 4T_f - 40 \] \[ 270 + 40 = 9T_f + 4T_f \] \[ 310 = 13T_f \] \[ T_f = \frac{310}{13} \approx 23.846^\circ C \] Step 4: Final Answer:
The temperature of the mixture when x and z are mixed is approximately 23.84\(^{\circ}\)C. This corresponds to option (B).