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प्रश्न
200 g of hot water at 80°C is added to 400 g of cold water at 10°C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg-1K-1.
उत्तर
Mass of hot water (m1) = 200g
Temperature of hot water (T1) = 80°C
Mass of cold water (m2) = 400g
Temperature of cold water (T2) =10°C
Final temperature (T) = ?
m1 c1 (T1 - T) = m2 c2 (T - T2)
c1 = c2
T = `("m"_1 "T"_1 + "m"_2 "T"_2)/("m"_2 + "m"_1)`
`T = (200 xx 80 + 400 xx 10)/600`
T = 33.3°C
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