A bucket contains 8 kg of water at 25°C and 2 kg of water at 80°C is poured into it. Neglecting the heat absorbed by the bucket, calculate the final temperature of water.
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Answered by
82
Specific heat capacity of water=4200 J/kgK
By the principle of mixtures,
m x c (T1-T) = m x c x(T-T2)
2 x 4200(80-T) = 8 x 4200(T-25)
(4200 on both sides get cancelled)
160-2T = 8T-200
360 = 10T
T= 36 degree Celcius
By the principle of mixtures,
m x c (T1-T) = m x c x(T-T2)
2 x 4200(80-T) = 8 x 4200(T-25)
(4200 on both sides get cancelled)
160-2T = 8T-200
360 = 10T
T= 36 degree Celcius
Answered by
35
let the final temperature be T
s be specific heat of water
amount of heat in bucket, hot water is equal to the final mixed water
8 *25 * s + 2 * 80 * s = (8+2) * s * T
T = 36 deg C
s be specific heat of water
amount of heat in bucket, hot water is equal to the final mixed water
8 *25 * s + 2 * 80 * s = (8+2) * s * T
T = 36 deg C
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