Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
1 kg of ice at – 10°C is mixed with 4.4 kg of water at 30°C. The final temperature of mixture is: (specific heat of ice is 2100 J/kg/k) . 1 kg of ice at – 10°C is mixed with 4.4 kg of water at 30°C. The final temperature of mixture is: (specific heat of ice is 2100 J/kg/k) .
Mass of ice, m = 1kg at -10°C mass of water, M = 4.4kg at 30°C heat flows through water to ice. so, heat gained by ice and heat lost by water. total heat gained by ice to melt = heat required by ice to reach the temperature 0°C + heat require to melt of ice = ms∆T + mL = 1 kg × 2100 J/Kg/°C × (0 -(-10)) + 1kg × 336000/Kg= 21000 + 336000 = 357000 J heat lost by water to reduce the temperature at 0°C = MS∆T'= 4.4kg × 4200 × (30 - 0) = 554400> 357000 so, temperature of mixture must be greater than zero. now, let's take, after melting ice reaches temperature A. then, heat is required to change temperature A is (554400 - 357000) = 197400J use formula, (m + M)S∆T = 197400 or, (1 + 4.4)× 4200 × (A - 0) = 197400A = 8.7°C
Dear student Check out this short method.Given that,Mass of ice = 1 kgTemperature of ice = 283 KMass of water = 4.4 kgTemperature of water = 303 KSpecific heat of ice = 2100 J/kg-KWe know that,Specific heat of water = 4185.5 J/kg-KNow, the final temperature of mixture is :Tf = miTiCpi + mwTwCpw /(miCpi + mwCpw)Put al the values Tf = 301k Hope this helps
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -