50 g of a solid at its freezing point -20°C is dropped into an ice-water mixture at 0°C. The substance melts and its liquid reaches 0°C with the whole mixture still at 0°C. For the substance L f = 3

Question

50 g of a solid at its freezing point -20°C is dropped into an ice-water mixture at 0°C. The substance melts and its liquid reaches 0°C with the whole mixture still at 0°C. For the substance $L_{f} = 3$ $cal / g$ and $c_{liquid} = 0.12 cal / (g^{\circ} C)$ ; for water $L_{f} = 80 cal / g$ . Find the additional ice formed, in g (three decimals).

Answer:

3.38

Solution:

The substance absorbs 50(3) = 150 cal to melt, then $50 (0.12) (20) = 120 cal$ to warm. Total 270 cal, which freezes $\frac{270}{80} = 3.375 g$ of water.

TestHub · Accepted Answer

The substance absorbs 50(3) = 150 cal to melt, then 50 ( 0.12 ) ( 20 ) = 120 cal to warm. Total 270 cal, which freezes 80 270 ​ = 3.375 g of water.

Physics - Heat & Thermo Question with Solution | TestHub

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