In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing $$150 cm^3$$ of water at 27 °C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for the specific heat of the metal?

Asked by Abhisek | 1 year ago |  134

##### Solution :-

Mass of the metal block, m = 0.20 kg = 200 g
Initial temperature of the metal block, T1​ = 150°C
Final temperature of the metal block, T2​ = 40°C
Copper calorimeter has water equivalent of mass, m1= 0.025 kg

= 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C is 150×1

=150g

Specific heat of water, Cw=4.186J/g/K
Specific heat of the metal = c

Decrease in the temperature of the metal block

ΔT1=T1​–T2​ =150−40=110°C
Increase in the temperature of the water and calorimeter system,

ΔT2=40−27=13°C

Heat lost by the metal

= Heat gained by the water + heat gained by the  calorimeter

mCΔT1 = (M+m1)CwΔT2

C =$$\dfrac{(M+m_1)C_wΔT_2}{mΔT_1}$$

$$\dfrac{(150 +25) \times 4.186 \times 13}{(200 \times 100)}$$

=$$\dfrac{(175 \times 4.186 \times 13)}{(200 \times 100)}$$

$$\dfrac{9523.15}{2200}$$

= 0.43 Jg−1k−1

Answered by Pragya Singh | 1 year ago

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