Question

1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to equilibrium at -8.88 degrees Celsius. What is the specific heat of the metal?

Answer 1

2178 J/(kg*C)

Step-by-step explanation:

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Answer 2

C₂ = 2.22 KJ/kg °C

Step-by-step explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

`Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_(1)C_(1) \Delta T_(1) = m_(2)C_(2) \Delta T_(2)`

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

`(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_(2)(13.88\ ^0C) \\C_(2) = (30.84\ KJ)/(13.88\ kg/^0C)`

C₂ = 2.22 KJ/kg °C