contestada

A student drops two metallic objects into a 120 g steel container holding) 150 g of water at 25°C. One object is a 253 g cube of copper that is initially at 85°C, and the other is a chunk of aluminum that is initially at 5°C. To the surprise of the student, the water reaches a final temperature of 25°C, its initial temperature. What is the mass of the aluminum chunk? (please add an explanation)

Respuesta :

fichoh

Answer: 0.326kg

Explanation:

Given the following :

Mass of copper (Mc) = 253g

Initial temperature of copper (t1) = 85°C

Mass of aluminium(Ma)=?

Initial temperature of aluminum(t1) = 5°C

Specific heat capacity of aluminum(Ca) = 900

Specific heat capacity of copper (Cc) = 387

Final temperature(t2) = 5°C

Q = mc(t2 - t1)

Ma × Ca × (t2 - t1)a = Mc × Cc × (t2 - t1)c

Ma × 900 × (25 - 5) = 0.253 × 387 × (25 - 85)

Ma × 900 × 20 = 0.253 × 387 × 60

Ma × 18000 = 5874.66

Ma = 5874.66 / 18000

Ma = 0.32637

Ma = 0.326kg

onyebuchinnaji

The mass of the aluminum chunk dropped into the water is 325.5 g.

The given parameters;

  • mass of the steel, = 120 g
  • mass of water = 150 g
  • initial temperature of water = 25 °C
  • mass of copper = 253 g
  • initial temperature of the copper = 85 °C
  • initial temperature of the aluminum = 5 °C
  • final temperature of water = 25 °C
  • Specific heat capacity of aluminum = 0.9 J/gC
  • Specific heat capacity of copper = 0.386 J/gC

Apply the principle of conservation of energy, the heat gained by the aluminum is equal to heat lost by the copper;

[tex]M_a C_a (t- t_a) = M_c C_c (t_c - t)\\\\M_a\times 0.9 \times (25-5) = 253 \times 0.386 \times (85-25)\\\\18 M_a = 5859.48\\\\M_a = \frac{5859.48}{18} \\\\M_a = 325 .5 \ g[/tex]

Thus, the mass of the aluminum chunk dropped into the water is 325.5 g.

Learn more here: https://brainly.com/question/20709851

Otras preguntas