JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
When \(100\,g\) of a liquid \(A\) at \(100\,^oC\) is added to \(50\,g\) of a liquid \(B\) at temperature \(75\,^oC\), the temperature of the mixture becomes \(90\,^oC\). The temperature of the mixture, if \(100\,g\) of liquid \(A\) at \(100\,^oC\) is added to \(50\,g\) of liquid \(B\) at \(50\,^oC\), will be ........\(^oC\)
- A \(85\)
- B \(60\)
- C \(80\)
- D \(70\)
Answer & Solution
Correct Answer
(C) \(80\)
Step-by-step Solution
Detailed explanation
\(100 \times {S_A} \times \left[ {100 - 90} \right] = 50 \times {S_B} \times \left( {90 - 75} \right)\) \(2{S_A} = 1.5\,{S_B}\) \({S_A} = \frac{3}{4}{S_B}\) \(Now,100 \times {S_A} \times \left[ {100 - T} \right] = 50 \times {S_B}\left( {T - 50} \right)\)…
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