TS EAMCET · Physics · Thermal Properties of Matter
The temperature of \(100 \mathrm{~g}\) of water is to be raised from \(24^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\) by adding steam at \(100^{\circ} \mathrm{C}\) to it. The mass of the steam required in this process is (latent heat of steam is \(540 \mathrm{cal} \mathrm{g}^{-1}\) )
- A \(2 \mathrm{~g}\)
- B \(4 \mathrm{~g}\)
- C \(10 \mathrm{~g}\)
- D \(12 \mathrm{~g}\)
Answer & Solution
Correct Answer
(D) \(12 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
(d) Apply principle of calorimetry \[ \begin{aligned} & 100 \times 1 \times(90-24)=540 m+m \times 1 \times(100-90) \\ & 6600=540 m+10 m \end{aligned} \] \(\begin{aligned} & 6600=550 \mathrm{~m} \\ & \Rightarrow \mathrm{m}=\frac{6600}{550}=12 \mathrm{~g}\end{aligned}\)
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