TS EAMCET · Physics · Thermal Properties of Matter
A thermos flask contains \(250 \mathrm{~g}\) of coffee at \(90^{\circ} \mathrm{C}\). To this \(20 \mathrm{~g}\) of milk at \(5^{\circ} \mathrm{C}\) is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as \(\left.1.00 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\right)\)
- A \(3.23^{\circ} \mathrm{C}\)
- B \(3.15^{\circ} \mathrm{C}\)
- C \(83.7^{\circ} \mathrm{C}\)
- D \(37.8^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(83.7^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Let after getting equilibrium stage final temperature is \(T\). Then, according to loss of calorimetry Heat loss \(=\) Heat gain \[ \begin{aligned} \Rightarrow(250)(1)(90-T) & =20(1)(T-5) \\ \Rightarrow \quad T & =83.7^{\circ} \mathrm{C} \end{aligned} \]
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