Question

ГОСПОДИ ЭТО СРОЧНООООООООО ПОМОГИТЕ РЕШИТЬ ПОЖАЛУЙСТА!!!!!!!!!!!!!!!!! МЕТАЛЛИЧЕСКУЮ ДЕТАЛЬ, МАССА КОТОРОЙ 200 Г, НАГРЕЛИ ДО 100ГРУДУС , А ЗАТЕМ ОТПУСТИЛИ В ВОДУ МАССОЙ 800 Г И ТЕМПЕРАТУРОЙ 20 ГРАДУС . ЧЕРЕЗ НЕКОТОРОЕ ВРЕМЯ ТЕМПЕРАТУРА ВОДЫ И ДЕТАЛИ СТАЛА РАВНА 25 ГРАДУСОВ. ОПРЕДЕЛИТЕ УДЕЛЬНУЮ ТЕПЛОЁМКОСТЬ МЕТАЛЛИЧЕСКОЙ ДЕТАЛИ. РЕБЯТ ЭТО ООООЧЕНЬ СРОЧНО УМОЛЯЮ ПОМОГИТЕ​

Answer 1

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Answer 2

Calculation of Specific Heat Capacity

To determine the specific heat capacity of the metallic object, we can use the formula:

Q = mcΔT

Where: - Q is the heat transferred - m is the mass of the object - c is the specific heat capacity of the object - ΔT is the change in temperature

In this case, the mass of the object is 200 g, the initial temperature is 100°C, and the final temperature is 25°C. The mass of the water is 800 g, and its initial temperature is 20°C.

To find the specific heat capacity of the metallic object, we need to calculate the heat transferred to the water and equate it to the heat transferred from the metallic object.

Calculation of Heat Transferred to Water

The heat transferred to the water can be calculated using the formula:

Q = mcΔT

Where: - Q is the heat transferred - m is the mass of the water - c is the specific heat capacity of water (which is approximately 4.18 J/g°C) - ΔT is the change in temperature

Substituting the given values, we have:

Q = (800 g)(4.18 J/g°C)(25°C - 20°C)

Calculating this expression, we find that the heat transferred to the water is approximately 836 J.

Equating Heat Transferred to Water and Heat Transferred from Metallic Object

Since the heat transferred to the water is equal to the heat transferred from the metallic object, we can set up the equation:

Q_water = Q_metal

Substituting the calculated values, we have:

836 J = (200 g)(c)(100°C - 25°C)

Simplifying this equation, we find:

836 J = 17500 c

Solving for c, the specific heat capacity of the metallic object, we divide both sides of the equation by 17500:

c ≈ 0.0477 J/g°C

Therefore, the specific heat capacity of the metallic object is approximately 0.0477 J/g°C.

Please note that this calculation assumes that the heat transfer is solely between the metallic object and the water, and that no heat is lost to the surroundings.