In 2 kg of water t 80 ° C, water t 10 ° C was added, resulting in a temperature of 35 ° C.
| August 14, 2021 | education
In 2 kg of water t 80 ° C, water t 10 ° C was added, resulting in a temperature of 35 ° C. What is the weight of the added water? Neglect the ambient temperature.
m1 = 2 kg.
t1 = 80 ° C.
t2 = 10 ° C.
t3 = 35 ° C.
Cw = 4200 J / kg * ºC.
m2 -?
When mixed, hot water, cooling from temperature t1 to t3, will give the amount of heat to cold water, which will be heated from temperature t2 to t3.
The heat balance equation will have the form: C * m1 * (t1 – t3) = C * m2 * (t3 – t2), where C is the specific heat capacity of water, m1, m2 is the mass of hot and cold water.
m2 = C * m1 * (t1 – t3) / C * (t3 – t2) = m1 * (t1 – t3) / (t3 – t2).
m2 = 2 kg * (80 ° C – 35 ° C) / (35 ° C – 10 ° C) = 3.6 kg.
Answer: cold water was added m2 = 3.6 kg.
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