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A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises from 22.5 to 25.0 ◦C. The specific heat of water is 4.18 Jg−1◦C −1 . Calculated the specific heat of the metal. Express your answer in Jg−1◦C −1 .

Respuesta :

Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

Step 2: Calculate the specific heat of the metal

Qgained  = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C

mickymike92

Based on the data provided, the specific heat capacity of the metal is 0.335 J/g°C

What is specific heat capacity of a substance?

The specific heat capacity, c of a substance is the amount of heat required to increase the temperature of a unit mass of the substance by 1 °C or 1 K.

The quantity of Heat, Q is given by the formula below:

[tex]Q= m*c*ΔT[/tex]

Where:

  • m is mass of substance
  • c is specific heat capacity
  • ΔT is temperature change

The specific heat capacity of the metal is calculated from the data provided:

  • mass of the metal = 12.0 grams
  • initial temperature of the metal = 90.0 °C
  • mass of the water = 25.0 grams
  • initial temperature of water = 22.5 °C
  • final temperature of water (and metal) = 25.0 °C
  • specific heat capacity of water = 4.18 J/g°C

Using the heat conservation principle;

  • heat lost = heat gained
  • Qgained  = -Qlost
  • Qwater = -Qmetal

-Qmetal = 12 × c × (25.0 - 90.0 °C)

Qmetal = 780c

Qwater = 25 × 4.18 × (25.0 - 22.5)

Qwater = 261.25

Thus;

c(metal) = 261.25/780

c(metal) = 0.335 J/g°C

Therefore, the specific heat capacity of the metal is 0.335 J/g°C.

Learn more about specific heat capacity at: https://brainly.com/question/26108629

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