What amount of water m at a temperature of t2 = 80 ° C must be added to m = 100 g of water at a temperature of t1 = 16 ° C
| December 12, 2020 | education
What amount of water m at a temperature of t2 = 80 ° C must be added to m = 100 g of water at a temperature of t1 = 16 ° C so that the final temperature of the mixture becomes t3 = 40 ° C?
Specific heat of water = c = 4200 J / (kg * K) Hot water gives off heat to cold water => Q1 = Q2 4200 * m * 40 = 4200 * 0.1 * 24 40 * m = 2.4 m = 0.06 ( kg)
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