contestada

Exactly 253.0 J will raise the temperature of 10.0 g of a metal from 25.0 °C to 60.0 °C. What is the specific heat capacity of the metal?

a. 0.723 J/(g. °C)

b. 1.38 J/(g°C)

C 12.2 J/g °C)

d. 60.5J/g - °C)

e. None of these

Respuesta :

tutorAnne

Answer:a. 0.723 J/(g. °C)

Explanation:

Q = mCΔT

where,

Q- Quantity of heat supplied-= 253.0j

m - mass of substance=10g

C - specific heat capacity=?

ΔT - change in temperature = 60 - 25 = 35°C

Imputing values, we have ,

Q = mCΔT

C = Q/(mΔT) = 253.0/ ( 10 x 35 ) = 0.7228 rounded to ---> 0.723 J/(g. °C)

Cricetus

The specific heat capacity of the metal will be "0.723 J/(g.°C)".

Heat and Temperature

According to the question,

Heat supplied quantity, Q = 253.0 J

Temperature, T₁ = 25.0°C

                       T₂ = 60.0°C

Change in temperature, ΔT = T₂ - T₁

                                               = 60 - 25

                                               = 35°C

We know the relation,

→ Q = mCΔT

or,

→ C = [tex]\frac{Q}{m \Delta T}[/tex]

By substituting the values, we get

      = [tex]\frac{253.0}{10\times 35}[/tex]

      = [tex]\frac{253.0}{350}[/tex]

      = 0.7228 or,

      = 0.723 J/(g.°C)

Thus the response above (Option a) is correct.

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https://brainly.com/question/26239739

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