Question

A metal initially at 450.℃ and mass 2.00 g is dropped into a liquid with specific heat capacity 4.00 Jg^-1℃^-1, mass 10.0 g and initial temperature 40.0℃. The final temperature of the liquid is 50.0℃. What is the specific heat capacity of the metal. Express your answer in Jg^-1℃^-1.

Answer 1

Answer:

0.5

Explanation:

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Answer 2

The specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹

SPECIFIC HEAT CAPACITY:

• The specific heat capacity of a metal can be calculated by using the following formula:

Q = mc∆T

Where;

1. Q = quantity of heat absorbed or released (J)
2. m = mass of substance (g)
3. c = specific heat capacity (Jg°C)
4. ∆T = temperature (°C)

• However, in a calorimeter, the following expression applies:

• mc∆T (liquid) = - (mc∆T) metal

METAL:

1. m = 2.00g
2. ∆T = 50°C - 450°C = -400°C
3. c = ?

LIQUID:

1. m = 10.0g
2. ∆T = 50°C - 40°C = 10°C
3. c = 4.00 Jg-¹℃-¹

• 10 × 4 × 10 = -{2 × -400 × c}

• 400 = -{-800c}

• 400 = 800c

• c = 400 ÷ 800

• c = 0.5Jg-¹°C-¹

Therefore, the specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹.

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