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4. Q 4.500g of water at 100°C is mixed with 300g of water at 30° C. What is the temperature of themixture? Take specific heat capacity of water as4.2 J/kg/K.Full Marks : 2 |
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Answer» Mass of hot water, m 1 =500gMass of cold water, m 2 =300gTemp. of hot water, t 1 =100 o CTemp. of cold water, t 2 =30 o CSp. heat of water, C=4.2Jg −1 o C −1 Let temp. of mixture be t o C. Then, Heat gained by cold water=m 2 ×C×(t−t 2 )According to the PRINCIPLE of CALORIMETRY, Heat lost = Heat gained500×4.2×(100−t)=300×4.2×(t−30)⇒5(100−t)=3(t−30)⇒−3t−5t=−90−500⇒−8t=−590⇒t= 8590 =73.8 o CSo, the final temperature of the mixture is 73.8 o C. |
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