300g of water at 25^○C is added to 100g 0f ice at 0^○C. The final temperature of mixture is

1. -5/2^○C
2. -5/3^○C
3. -5^○C
4. 0^○C

The answer is 0^○ but I got -5/4 using Q1 + Q2=Q3 formula PLEASE HELP!!!!

1 st find heat content of water by Q=mst And latent heat req. To melt ice by Q=ml ... U will see that water is not able to melt the entire ice.... Therefore some ice will remain and temp. Will be 0 Degree Celsius

Dear Student

The heat required for 100g of ice at 0^∘C to change into
Water at0^∘C = mL = 100×80×4.2 = 33,600J…(i)
The heat released by 300g of water at 25^∘C to change its
temperature to 0^∘C = mcΔT = 300×4.2×25 = 31,500J….(ii)
Since the energy in eq.(ii) is less than of eq(i)
therefore the final temperature will be 0^∘C

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya