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 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.09

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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
$∴ 900 \times 1 \times (20 - 19) = 1000 \times s (19 - 10)$
$\Rightarrow s = 0.1 C G S U n i t s .$

Suggest Corrections

Similar questions

Q. A calorimeter of mass

50 g

and specific heat capacity

0.42 J g^{- 1} ℃^{- 1}

contains some mass of water at

20^{o} C

. A metal piece of mass

20 g

100^{o} C

is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be

22^{o} C

. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece

= 0.3 J g^{- 1} ℃^{- 1}

specific heat capacity of water

= 4.2 J g^{- 1} ℃^{- 1}

]

Q. A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J kg⁻¹ °C⁻¹.

A calorimeter contains 5 kg of water at 30°C. A metal of mass 4 kg at 100^oC is added to the calorimeter. The final temperature is recorded as 70^oC. Find the specific heat capacity of the metal. Given: Specific heat capacity of water 4200 J kg^–1°C^–1.

A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at $20^{\circ} C$ . The mixture attains an equilibrium teperature of $60^{\circ} C .$ Find the temperature of the furnace. Specific heat capacity of iron $= 470 H k g^{- 1}^{\circ} C^{- 1} .$

Q. If A vessel of thermal capacity

30

calories contains

170 g

of water at

30^{\circ} C

. If on dropping a solid at

93^{\circ} C

into the calorimeter, then temperature of mixture becomes

33^{\circ} C

, find the thermal capacity of solid.

Join BYJU'S Learning Program

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.