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

Answer:

2178 J/(kg*C)

Explanation:

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Answer 2

Answer:

C₂ = 2.22 KJ/kg °C

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:

$(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}$

C₂ = 2.22 KJ/kg °C