Download the app

Calorimetry

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept
By Expert Faculty of Sri Chaitanya
Question

2 kg of ice at -20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel in kg. It is given that the specific heat of water and ice are 1 kcal/kg/°C and 0.5 kcal/kg/°C respectively and the latent heat of fusion of ice is 80 kcal/kg.

Moderate
Solution

Let m kg be the mass of ice melted into water. 

Heat lost by 5 kg of water = 5 kg × 1 kcal/kg/°C × 20°C = 100 kcal. 

Heat gained = mL+msΔT=m kg × 80 kcal/kg + 2 kg × 0.5 kcal/kg/°C × 20°C = 80 m kcal + 20 kcal.

 Now, heat gained = heat lost. 

Therefore, 80 m + 20 = 100

or m=1 kg. 

Therefore, final mass of water = 5 kg + 1 kg = 6 kg.


NEW

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


ctaimg

Create Your Own Test
Your Topic, Your Difficulty, Your Pace


Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as S=1+θ500 cal g1C1where θ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are Sice =0.5 cal g1C1Swater =1.0 cal g1C1 and latent heat of fusion of ice is Lf=80calg1. Assume no heat loss to the surrounding and calculate the value of m in grams. 

Get your all doubts cleared from
Sri Chaitanya experts

counselling
india
+91

phone iconwhats app icon