You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Specific Heat Capacity

A hot solid o...

Question

A hot solid of mass $60 g$ at $100^{o} C$ is placed in 150 g of water at $20^{o} C$ . The final steady temperature recorded is $25^{o} C$ . Calculate the specific heat capacity of the solid.
[Specific heat capacity of water = 4200 J $k g^{- 1}^{o} C^{- 1}$ ].

Open in App

Solution

Heat gained $=$ Heat lost
$150 \times 4.4 \times (25 - 50) = 60 \times c \times (100 - 25)$
$150 \times 4.2 \times 5 = 60 \times c \times 75$
Specific heat capacity $(c) = 0.7 J g^{- 1}^{o} C^{- 1}$

Suggest Corrections

Similar questions

Q. A solid of mass

50

g at

150^{o}

C is placed in

100

g of water at

11^{o}

C, when the final temperature recorded is

20^{o}

C. Find the specific heat capacity of the solid. (Specific heat capacity of water

4.2

/ g^{o}

A hot solid of mass 60 g at 100 $^{\circ} C$ is placed in 150 g of water at 20 $^{\circ} C$ . The final temperature recorded is 25 $^{\circ} C$ . Calculate the specific heat of solid. (Specific heat of water = 4200J/kg/ $^{\circ} C$ )

Q. A solid of mass 50g at

150^{\circ} C

is placed in 100 g of water at

11^{\circ} C

when the final temperature recorded is

20^{\circ} C

. Find the specific heat capacity of solid. SHC of water is

4.2 J / g^{\circ} C

Q. 200 g of hot water at

80^{0}

C is added to 300 g of cold water at

10^{0}

C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water . Specific heat capacity of water = 4200 J

k g^{- 1} K^{- 1}

Q. The amount of heat energy required to convert

1 k g

of ice at

- 10^{o} C

to water at

100^{o} C

7, 77, 000 J

. Calculate the specific latent heat of ice. Specific heat capacity of ice

= 2100 J k g^{- 1} K^{- 1}

, specific heat capacity of water

= 4200 J k g^{- 1} K^{- 1}

Join BYJU'S Learning Program

A hot solid of mass 60g at 100oC is placed in 150 g of water at 20oC. The final steady temperature recorded is 25oC. Calculate the specific heat capacity of the solid.[Specific heat capacity of water = 4200 J kg−1oC−1].

Heat gained = Heat lost150×4.4×(25−50)=60×c×(100−25)150×4.2×5=60×c×75Specific heat capacity (c)=0.7Jg−1oC−1

A hot solid of mass $60 g$ at $100^{o} C$ is placed in 150 g of water at $20^{o} C$ . The final steady temperature recorded is $25^{o} C$ . Calculate the specific heat capacity of the solid.
[Specific heat capacity of water = 4200 J $k g^{- 1}^{o} C^{- 1}$ ].

Heat gained $=$ Heat lost
$150 \times 4.4 \times (25 - 50) = 60 \times c \times (100 - 25)$
$150 \times 4.2 \times 5 = 60 \times c \times 75$
Specific heat capacity $(c) = 0.7 J g^{- 1}^{o} C^{- 1}$