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A 21.3 g sample of a metal is heated to 70.0°C and dropped into 62.4 g of the water at 21.0°C. The final temperature of the water is 24.0°C. Given that the specific heat capacity of water is 4.184 Jg-1°C-1 calculate the specific heat of the metal.

User Xiaokun
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To solve for the specific heat of the metal, we can use the equation:

heat lost by metal = heat gained by water

The heat lost by the metal can be calculated using the equation:

Q = m × c × ΔT

where:
m = mass of the metal (21.3 g)
c = specific heat of the metal (unknown)
ΔT = change in temperature of the metal (final temperature of water - initial temperature of metal)

ΔT = 24.0°C - 70.0°C = -46.0°C

Q = 21.3 g × c × (-46.0°C)

The heat gained by the water can be calculated using the equation:

Q = m × c × ΔT

where:
m = mass of the water (62.4 g)
c = specific heat of water (4.184 Jg^-1°C^-1)
ΔT = change in temperature of the water (final temperature - initial temperature)

ΔT = 24.0°C - 21.0°C = 3.0°C

Q = 62.4 g × 4.184 Jg^-1°C^-1 × 3.0°C

Since the heat lost by the metal is equal to the heat gained by the water, we can equate the two equations:

21.3 g × c × (-46.0°C) = 62.4 g × 4.184 Jg^-1°C^-1 × 3.0°C

Solving for c:

c = [62.4 g × 4.184 Jg^-1°C^-1 × 3.0°C] / [21.3 g × (-46.0°C)]

c ≈ 0.38 Jg^-1°C^-1

Therefore, the specific heat of the metal is approximately 0.38 Jg^-1°C^-1.
User Justine Krejcha
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