KCET · Physics · Thermal Properties of Matter
\(100 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(100 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\). The final temperature of the mixture is
[Take, \(L_f=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\) and \(\left.S_w=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right]\)
- A \(40^{\circ} \mathrm{C}\)
- B \(10^{\circ} \mathrm{C}\)
- C \(50^{\circ} \mathrm{C}\)
- D \(1^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(10^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Let the final temperature of mixture be \(T\).
Heat absorbed by ice \(=\) heat lost by water
\(\therefore m L+m c(T-0)=m c(100-T)\)
Substituting values,
\(\therefore (100 \times 80)+(100 \times 1) T=100 \times 1(100-T) \)
\( 8000+100 T=10000-100 T \)
\( 200 T=2000 \)
\( T=10^{\circ} \mathrm{C} \)
Heat absorbed by ice \(=\) heat lost by water
\(\therefore m L+m c(T-0)=m c(100-T)\)
Substituting values,
\(\therefore (100 \times 80)+(100 \times 1) T=100 \times 1(100-T) \)
\( 8000+100 T=10000-100 T \)
\( 200 T=2000 \)
\( T=10^{\circ} \mathrm{C} \)
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