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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
A calorimeter of water equivalent \(20\, g\) contains \(180\, g\) of water at \(25^{\circ} C\). '\(m\)' grams of steam at \(100^{\circ} C\) is mixed in it till the temperature of the mixure is \(31^{\circ} C\). The value of \('m'\) is close to (Latent heat of water \(=540\) cal \(g ^{-1}\), specific heat of water \(=1\) cal \(g^{-1}{ }^{\circ} C ^{-1}\) )
- A \(2.6\)
- B \(2\)
- C \(4\)
- D \(3.2\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(\frac{ Cal }{20 gm } \frac{ H _{2} O }{180 gm } \quad \frac{ Steam }{ m }\) \(200 \times 1 \times(31-25)\)\(= m \times 540+ m \times 1 \times(100-31)\) \(m \approx 2 \;gm\)
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