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A cool water sample absorbed 3,135 J of energy from hot metal. The temperature of the 63.0 g piece of metal changed from 382°C to 20.0°C.

What is the specific heat of the metal?
Cₘₑₜₐₗ=[?] J/g·°C
qH₂O=3,135J

User Khonix
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1 Answer

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Final answer:

To calculate the specific heat of the metal, use the heat transfer formula and the given values: q = 3,135 J, m = 63.0 g, and ΔT = 362°C. Solving for the specific heat capacity, c is determined to be 0.138 J/g°C.

Step-by-step explanation:

To determine the specific heat of the metal, we can use the formula:

q = m × c × ΔT

where:

  • q is the heat absorbed or released (in joules),
  • m is the mass of the substance (in grams),
  • c is the specific heat capacity (in J/g°C), and
  • ΔT is the change in temperature (in °C).

The question states that the metal has released 3,135 J of energy as it cooled down, which is the same amount of energy absorbed by the water. We're given that the mass (m) of the metal is 63.0 g and the temperature change (ΔT) is from 382°C to 20.0°C, thus ΔT is 382 - 20 = 362°C.

Plugging these values into the formula, we have:

3,135 J = 63.0 g × c × 362°C

To find c, the specific heat, rearrange the formula:

c = 3,135 J / (63.0 g × 362°C)

c = 0.138 J/g°C

So, the specific heat capacity of the metal is 0.138 J/g°C.

User Todd Lehman
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