A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25°C is cooled by adding 20 gm ice at 0°C. cIf all the ice melts, the final temperature of the drink is : (density of drink = 0.833 gm/cc, specific heat of alcohol = 0.6 cal/gm/°C, specific heat of water = cal/gm/°C, latent heat of fusion of ice = 80cal/gm)
Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.
Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA
mdrink = 0.833 × 120 = 100g (50g water + 50g Alcohol)
Heat given by ice = Heat taken by drink
mice ℓice + mice Swater (T – 0) = mAℓ SAℓ (25 – t) + mwater Swater (25 – t)
mice = 20g ℓice = 80 cal/g
mAℓ = 50g SAℓ = 0.6 cal/g
T = 4°C