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

Concept: Heat transfer involves: • Sensible heat: \(Q = mc\Delta T\) • Latent heat: \(Q = mL\) Given: • Specific heat of water \(c = 1\,\text{cal/g}^\circ\text{C}\) • Latent heat of vaporisation \(L = 540\,\text{cal/g}\)
Step 1: Heat released when \(x\) gm cools from \(50^\circ\text{C}\) to \(0^\circ\text{C}\). \[ H = mc\Delta T \] \[ H = x(1)(50 - 0) \] \[ H = 50x \qquad ...(1) \]
Step 2: Heat required to convert \((100 - x)\) gm water at \(50^\circ\text{C}\) to steam at \(100^\circ\text{C}\). Two steps involved: • Heating from \(50^\circ\text{C}\) to \(100^\circ\text{C}\) • Vaporisation at \(100^\circ\text{C}\) \[ H = (100 - x)(1)(100 - 50) + (100 - x)(540) \] \[ H = (100 - x)(50 + 540) \] \[ H = (100 - x)(590) \qquad ...(2) \]
Step 3: Equate the two heats. \[ 50x = (100 - x)(590) \] \[ 50x = 59000 - 590x \] \[ 640x = 59000 \] \[ x = \frac{59000}{640} \] \[ x = 92.18\,\text{gm} \] Thus, \[ \boxed{x = 92.18\,\text{gm}} \]