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The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). What is the final temperature if 305 J of heat is added to 71.6 g of this metal, initially at 20.0 ∘C?

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Final answer:

The final temperature is 23.89 °C.

Step-by-step explanation:

The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). To find the final temperature when 305 J of heat is added to 71.6 g of this metal initially at 20.0 ∘C, we can use the specific heat formula:

q = m * c * ΔT

Where:

  • q is the heat added
  • m is the mass of the metal
  • c is the specific heat of the metal
  • ΔT is the change in temperature

Substituting the known values into the formula:

305 J = (71.6 g) * (0.128 J/(g⋅∘C)) * ΔT

Solving for ΔT:

ΔT = 305 J / (71.6 g * 0.128 J/(g⋅∘C))

ΔT ≈ 3.89 ∘C

The final temperature is 20.0 ∘C + 3.89 ∘C = 23.89 ∘C

User Danixa
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5 votes

Final answer:

To find the final temperature of the metal when 305 J of heat is added to 71.6 g of it initially at 20.0 °C, we use the specific heat equation. The change in temperature (ΔT) is calculated to be 33.2 °C, resulting in a final temperature of 53.2 °C.

Step-by-step explanation:

The question involves the concept of specific heat capacity in physics, which is the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. The specific heat capacity formula is given by:

q = m × c × ΔT

where q is the amount of heat added (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).

Given that the specific heat of the metal is 0.128 J/(g°C), the mass of the metal is 71.6 g, the initial temperature is 20.0 °C, and the heat added is 305 J, we can rearrange the above formula to solve for the final temperature:

ΔT = °C = q / (m × c)

Plugging in the values gives us ΔT = 305 J / (71.6 g × 0.128 J/(g°C))

ΔT = 33.2 °C (rounded to one decimal place)

To find the final temperature, we add the change in temperature to the initial temperature:

T_final = T_initial + ΔT = 20.0 °C + 33.2 °C

T_final = 53.2 °C

User Marco Craveiro
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7.7k points
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