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Standard XII

Physics

Calorimetry

If 10 g of li...

Question

If 10 g of liquid at 300K is heated to 350 K, the liquid absorbs 6 kcals. Determine the specific heat of the liquid in cal/ $g^{\circ} C$ .

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120

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600

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Solution

The correct option is D 12
Specific heat = $\frac{e n e r g y a b s o r b e d}{s a m p l e m a s s \times T e m p e r a t u r e c h a n g e} = \frac{6000}{10 \times 50} = 12 c a l / g$

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Similar questions

10 g

of liquid at

300 K

is heated to

350 K

. The liquid absorbs

6 k c a l s

. What is the specific heat of the liquid (in

c a l / g^{o} C

Property	Value
Heat of Fusion (cal/g)	80
Heat of Vaporization (cal/g)	540
Specific Heat (cal/g. $^{o}$ C)	1.00

For conversion of 10 grams of ice at 0

^{o}

C to liquid water at 100

^{o}

C heat needed is:

Q. When a block of metal of specific heat

0.1 c a l / g^{\circ} C

g and weighing

110 g

is heated to

100^{\circ} C

and then quickly transferred to a calorimeter containing

200 g

of a liquid at

10^{\circ} C

, the resulting temperature is

18^{\circ} C

. On repeating the experiment with

400 g

of same liquid in the same calorimeter at same initial temperature, the resulting temperature is

{14.5}^{\circ} C

. If the specific heat of the liquid is

\frac{x}{100} c a l / g^{\circ} C

. Find '

x

Q. A liquid of specific heat

0.3 c a l / g^{o} C

90^{o} C

is mixed with another liquid of specific heat

0.5 c a l / g^{o} C

15^{o} C

. If the resultant temperature of mixture is

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then the ratio of their masses is:

Q. How many grams of a liquid of specific heat

0.2 {cal/g}^{\circ} C

at a temperature

40^{\circ} C

must be mixed with

100 g

of a liquid of specific heat

0.5 {cal/g}^{\circ} C

at a temperature

20^{\circ} C

, so that the final temperature of the mixture becomes

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If 10 g of liquid at 300K is heated to 350 K, the liquid absorbs 6 kcals. Determine the specific heat of the liquid in cal/g∘C.

The correct option is D 12Specific heat = energy absorbedsample mass×Temperature change=600010×50=12cal/g

If 10 g of liquid at 300K is heated to 350 K, the liquid absorbs 6 kcals. Determine the specific heat of the liquid in cal/ $g^{\circ} C$ .

The correct option is D 12
Specific heat = $\frac{e n e r g y a b s o r b e d}{s a m p l e m a s s \times T e m p e r a t u r e c h a n g e} = \frac{6000}{10 \times 50} = 12 c a l / g$