Question

(3 бали) Розв'яжіть задачу. У холодну воду масою б кг, температура якої 15 °C, долили 4 кг гарячої води. Якою була температура гарячої води, якщо температура суміші виявилася такою, що дорівнює 45 C2 Питома теплоємність води 4200 Дж/(кг. "С). Пожалуста срочно нада

Answer 1

Відповідь:

t3=42⁰

t1=15⁰

m1=6кг

m2=4кг

t2-?

m3(води)=m1+m2=6+4=10кг

Q1=m1*c*t1=6*4200*15=378000Дж

Q3=m3*c*t3=10*4200*42=1764000Дж

t2=(Q3-Q1)/c*m2=(1764000-378000)/4200*4=82,5⁰

Пояснення:

Answer 2

Problem Analysis

We are given that a certain amount of cold water with a mass of b kg and a temperature of 15 °C is mixed with 4 kg of hot water. We need to find the temperature of the hot water if the resulting mixture has a temperature of 45 °C. The specific heat capacity of water is given as 4200 J/(kg·°C).

Solution

To solve this problem, we can use the principle of conservation of energy. The total heat gained by the cold water and the hot water should be equal to the total heat lost by the resulting mixture.

Let's denote the initial temperature of the hot water as T1 (unknown), the final temperature of the mixture as T2 (given as 45 °C), and the specific heat capacity of water as c (given as 4200 J/(kg·°C)).

The heat gained by the cold water can be calculated using the formula:

Q1 = m1 * c * (T2 - T1)

where: - Q1 is the heat gained by the cold water, - m1 is the mass of the cold water (b kg), - c is the specific heat capacity of water (4200 J/(kg·°C)), - T2 is the final temperature of the mixture (45 °C), - T1 is the initial temperature of the hot water (unknown).

The heat gained by the hot water can be calculated using the formula:

Q2 = m2 * c * (T2 - T1)

where: - Q2 is the heat gained by the hot water, - m2 is the mass of the hot water (4 kg), - c is the specific heat capacity of water (4200 J/(kg·°C)), - T2 is the final temperature of the mixture (45 °C), - T1 is the initial temperature of the hot water (unknown).

Since the total heat gained by the cold water and the hot water should be equal to the total heat lost by the resulting mixture, we can set up the equation:

Q1 + Q2 = 0

Substituting the formulas for Q1 and Q2, we get:

m1 * c * (T2 - T1) + m2 * c * (T2 - T1) = 0

Simplifying the equation, we have:

b * c * (T2 - T1) + 4 * c * (T2 - T1) = 0

Factoring out (T2 - T1), we get:

(b + 4) * c * (T2 - T1) = 0

Since (T2 - T1) cannot be zero (as the temperatures are different), we can divide both sides of the equation by (b + 4) * c:

T2 - T1 = 0 / ((b + 4) * c)

Simplifying further, we have:

T2 - T1 = 0

Therefore, the initial temperature of the hot water (T1) is equal to the final temperature of the mixture (T2), which is 45 °C.

Answer

The initial temperature of the hot water was also 45 °C.