When 100 g of boiling water at 100- C is added into a calorimeter containing 300 g of cold water at 10- C- temperature of the mixture becomes 20- C- Then a metallic block of mass 1 kg at 10- C is dipped into the mixture in the calorimeter- After reaching thermal equilibrium- the final temperature becomes 19- C- What is the specific heat of the metal in C-G-S- units-A- 0-01B- 0-3C- 0-09D- 0-1

The correct option is D 0.1Initially Heat lost by boiling water = Heat gained by cold water 100 × 1 × 80 = (300+w) 1 × 10 Where w is the water equilvalent of c ...

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Standard XII

Chemistry

First Law of Thermodynamics

When 100 g of...

Question

When 100 g of boiling water at $100^{\circ} C$ is added into a calorimeter containing 300 g of cold water at $10^{\circ} C$ , temperature of the mixture becomes $20^{\circ} C$ . Then a metallic block of mass 1 kg at $10^{\circ} C$ is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes $19^{\circ} C$ . What is the specific heat of the metal in C.G.S. units?

0.01

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0.1

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Solution

The correct option is D 0.1
Initially
Heat lost by boiling water = Heat gained by cold water
$100 \times 1 \times 80 = (300 + w) 1 \times 10$
Where w is the water equilvalent of calorimeter
$\Rightarrow w = 500 g$
When block is added
Total mass of water =100+500+300 = 900 g
$∴ 900 \times 1 \times (20 - 19) = 1000 \times s (19 - 10)$
$\Rightarrow s = 0.1 C G S U n i t s .$

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The correct option is D 0.1Initially Heat lost by boiling water = Heat gained by cold water 100×1×80=(300+w)1×10 Where w is the water equilvalent of calorimeter ⇒w=500 g When block is added Total mass of water =100+500+300 = 900 g ∴900×1×(20−19)=1000×s(19−10) ⇒s=0.1 CGS Units.