Question

10 gm of ice at –20⁰C is added to 10gm of water at 50⁰C. S_water = 1 cal/gm⁰C, S_ice = 0.5 cal/gm⁰C. $L_{f_{ice}}$= 80 cal/gm.

The resulting temperature is –

• A. –20⁰C
• B. 15⁰C
• C. 0⁰C
• D. 50⁰C

Answer 1

Answer: (C)

0⁰C

Answer 2

Some ice will left in the mixture.