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200 g of hot water at 80 C is added to 300 g of cold water at 10 C. calculate the final temperature of the mixture of water.

Calorimetry numerical Question: 11 [reference: Selina class 10 ICSE book exercise]

200 g of hot water at 80o C is added to 300 g of cold water at 10oC. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg-1 K-1.

Solution:

Mass of hot water (m1) = 200 g

Temperature of hot water (T1) = 800 C

Mass of cold water (m2) = 300 g

Temperature of cold water (T2) = 100 C

Final temperature (T) =?

m1c1 (T1 – T) = m2c2 (T – T2) …………… [1]

Here, c1 = c2

So. from [1]

T =[m1T1 + m2T2 ]/ [m2 + m1]

T = [(200 × 80) + (300 × 10)] / 500

T = (16000 + 3000) / 500

T = 19000 / 500

T = 38o C

See also  A mass of 50 g of a certain metal at 150° C is immersed in 100 g of water at 11° C. The final temperature is 20° C. Calculate the specific heat capacity of the metal.
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