What mass of cold water at a temperature of t1 = 13 must be poured into hot water with a volume of V2 = 2
What mass of cold water at a temperature of t1 = 13 must be poured into hot water with a volume of V2 = 2 liters at a temperature of t2 = 85 to obtain water at a temperature of t = 25.
Initial data: t1 (cold water temperature) = 13 ºС; V2 (hot water volume) = 2 l (m2 = 2 kg); t2 (hot water temperature) = 85 ºС, t3 (steady-state temperature) = 25 ºС.
Since the equilibrium temperature has been established, the ratio is correct:
Q1 = Q2: C * m1 * (t3 – t1) = C * m2 * (t2 – t3).
m1 * (t3 – t1) = m2 * (t2 – t3).
m1 = m2 * (t2 – t3) / (t3 – t1).
Let’s do the calculation:
m1 = 2 * (85 – 25) / (25 – 13) = 2 * 60/12 = 10 kg (10 l).
Answer: It is necessary to pour 10 liters of cold water into hot water. Initial data: t1 (cold water temperature) = 13 ºС; V2 (hot water volume) = 2 l (m2 = 2 kg); t2 (hot water temperature) = 85 ºС, t3 (steady-state temperature) = 25 ºС.
Since the equilibrium temperature has been established, the ratio is correct:
Q1 = Q2: C * m1 * (t3 – t1) = C * m2 * (t2 – t3).
m1 * (t3 – t1) = m2 * (t2 – t3).
m1 = m2 * (t2 – t3) / (t3 – t1).
Let’s do the calculation:
m1 = 2 * (85 – 25) / (25 – 13) = 2 * 60/12 = 10 kg (10 l).
Answer: Pour 10 liters of cold water to hot water.
