Question

If \(x\) gm of water at \(50^\circ\text{C}\) cools to \(0^\circ\text{C}\), the heat released is \(H\). If \((100 - x)\) gm of water at \(50^\circ\text{C}\) is completely vapourised, the heat required is also \(H\). Find \(x\).

• A. \(8\,\text{gm}\)
• B. \(20\,\text{gm}\)
• C. \(92.18\,\text{gm}\)
• D. \(80\,\text{gm}\)

Hint:

When phase change is involved, always split the process: \[ \text{Total Heat} = mc\Delta T + mL \] First account for temperature change, then add latent heat if phase change occurs.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem relates to the principles of calorimetry.
It equates the heat released when a certain mass of water cools down to the heat required to heat up and vaporize another mass of water.
We will use the specific heat formula and the latent heat formula.
Step 2: Key Formula or Approach:
For cooling of water, the heat exchanged is $Q = ms\Delta T$.
For heating and phase change (vaporization), the heat exchanged is $Q = ms\Delta T + mL_v$.
Given that $H_1 = H_2 = H$.
Step 3: Detailed Explanation:
Calculate the heat released $H$ for the first mass:
\[ H = x \times 1 \times (50 - 0) \]
\[ H = 50x \quad \dots \text{(1)} \]
Calculate the heat required for the second process:
The $(100 - x)$ gm water first heats from 50$^\circ$C to 100$^\circ$C, and then changes phase to steam.
\[ H = (100 - x) \times 1 \times (100 - 50) + (100 - x) \times 540 \]
\[ H = (100 - x)(50) + (100 - x)(540) \]
\[ H = (100 - x)(590) \quad \dots \text{(2)} \]
Equating (1) and (2) since the heat amounts are equal:
\[ 50x = 59000 - 590x \]
\[ 640x = 59000 \]
\[ x = \frac{59000}{640} = 92.1875 \text{ gm} \]
Step 4: Final Answer:
The value of x is approximately 92.18 gm.