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. unit?

Question

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. unit? (A) 0.01 (B) 0.3 (C) 0.09 (D) 0.1

Tardigrade Admin · Accepted Answer

Let, heat capacity of calorimeter = ms 100 × 1 × 80=300 × 1 × 10+ ms × 10 ∴ ms =500 Cal / ° C Again, (100+300) × 1 × 1+500 × 1 =1000 × Sb × 9 ∴ Sb=0.1 Cal / g ° C

$0.01$

$0.3$

$0.09$

$0.1$

Let, heat capacity of calorimeter $= m s$
$100 \times 1 \times 80 = 300 \times 1 \times 10 + m s \times 10$
$∴ m s = 500 C a l /^{\circ} C$
Again, $(100 + 300) \times 1 \times 1 + 500 \times 1$
$= 1000 \times S_{b} \times 9$
$∴ S_{b} = 0.1 C a l / g^{\circ} C$

0.01

0.3

0.09

0.1

Let, heat capacity of calorimeter =ms 100×1×80=300×1×10+ms×10 ∴ms=500Cal/∘C Again, (100+300)×1×1+500×1 =1000×Sb​×9 ∴Sb​=0.1Cal/g∘C

$0.01$

$0.3$

$0.09$

$0.1$

Let, heat capacity of calorimeter $= m s$
$100 \times 1 \times 80 = 300 \times 1 \times 10 + m s \times 10$
$∴ m s = 500 C a l /^{\circ} C$
Again, $(100 + 300) \times 1 \times 1 + 500 \times 1$
$= 1000 \times S_{b} \times 9$
$∴ S_{b} = 0.1 C a l / g^{\circ} C$