COMEDK · Physics · 12. Thermal Properties of Matter
100 g of ice at \(0^{\circ} \mathrm{C}\) is mixed with 100 g of water at \(80^{\circ} \mathrm{C}\). The final temperature of the mixture is (Given, latent heat of fusion of ice, \(L_f=80 \mathrm{cal} / \mathrm{g}\) and specific heat capacity of water, \(S_{\text {water }}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) )
- A \(0^{\circ} \mathrm{C}\)
- B \(20^{\circ} \mathrm{C}\)
- C \(40^{\circ} \mathrm{C}\)
- D \(60^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(0^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Let the final temperature of mixture be \(T^{\circ} \mathrm{C}\). From principle of calorimetry, Heat given by 100 g of water \(=\) Heat taken by 100 g of ice \(m_1 S_{\text {water }}\left(\Delta T_1\right)=m_2 L_f+m_2 S_{\text {water }} \Delta T_2\) Substituting the given…
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