Question

500 calories are added to 100 gram copper at 25 degree celsius
a.what will be the final temperature?
b.if the same amount of heat was added to an equal amount of water,what would be the final temperature?​

Answer 1

Answer:

q = mass Cu x specific heat Cu x (Tfinal-Tinitial)

Substitute and solve for Tfinal.

b. Same for H2O but specific heat is not the same.

Explanation:

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Answer 2

(a.) The final temperature of the Copper would be 78.85°C and (b.) the final temperature of the water would be 30°C.

Given:

Amount of heat supplied (Q) = 500 calorie

Mass of copper present (m) = 100 gram

The temperature of copper (T₁) = 25°C

To Find:

(a.) Final temperature of Copper

(b.) The final temperature of the Water

Solution:

→ When heat energy is supplied to a certain substance then the temperature of the substance starts rising. The relationship between the amount of heat supplied and the temperature change is different for every material and it is described by the formula:

                                             Q = mcΔT

                            where Q = Amount of heat supplied

                                        m = mass of the substance

                                         c = specific heat capacity of the substance

                                      ΔT = Change in Temperature

(a.) Amount of heat supplied (Q) = 500 calorie = 4.2 × 500 = 2100 Joule

Mass of Copper present (m) = 100 gram = 0.1 kg

The initial temperature of copper (T₁) = 25°C

Let the final temperature of copper be 'T'.

Specific heat capacity of copper (c) = 390 J K⁻¹ kg⁻¹

∵ Q = mcΔT

∴ 2100 = 0.1 × 390 × (T-T₁)

→ 2100 = 39 × (T-25°)

→ 53.85 = T - 25°

    ∴ T = 78.85°C

So the final temperature of Copper would be 78.85°C.

(b.) Amount of heat supplied (Q) = 500 calorie = 4.2 × 500 = 2100 Joule

Mass of water present (m) = 100 gram = 0.1 kg

The initial temperature of water (T₁) = 25°C

Let the final temperature of water be 'T'.

Specific heat capacity of water (c) = 4200 J K⁻¹ kg⁻¹

∵ Q = mcΔT

∴ 2100 = 0.1 × 4200 × (T-T₁)

→ 2100 = 420 × (T-25°)

→ 5 = T - 25°

    ∴ T = 30°C

So the final temperature of the water would be 30°C.

Hence (a.) the final temperature of Copper would be 78.85°C and (b.) the final temperature of the water would be 30°C.

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