Question

У холодну воду масою 8 кг, температура 5 °С, долили 6 кг гарячої води. Якою була температура гарячої води, якщо температура суміші стала 35 °С. ( нада очень срочно!!! и подробно)поэтому даю много баллов

Answer 1

Дано m1=8 кг    t1=5C    t=35C    m2=6 кг   t2- ?

так как потерь тепла нет
Q2=Q1
c*m2*(t2-t)=c*m1*(t-t1)
m2*(t2-t)=m1*(t-t1)
6*(t2-35)=8*(35-5)
3*t2-105=120
3*t2=225
t2=75 С

Answer 2

Calculation of the Final Temperature

To determine the temperature of the hot water added to the cold water, we can use the principle of conservation of energy. The total heat gained by the cold water and the hot water is equal to the heat lost by the hot water. We can express this relationship using the equation:

m1 * c1 * (Tf - T1) = m2 * c2 * (T2 - Tf)

Where: - m1 is the mass of the cold water (8 kg) - c1 is the specific heat capacity of water (approximately 4.18 J/g°C) - Tf is the final temperature of the mixture (35°C) - T1 is the initial temperature of the cold water (5°C) - m2 is the mass of the hot water added (6 kg) - c2 is the specific heat capacity of water (approximately 4.18 J/g°C) - T2 is the initial temperature of the hot water (unknown)

By rearranging the equation, we can solve for T2:

T2 = (m1 * c1 * Tf + m1 * c1 * T1 - m2 * c2 * Tf) / (m2 * c2)