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Byju's Answer

Standard VII

Physics

Inter-Conversion of Temperature in Different Scales

A calorimeter...

Question

A calorimeter has mass $100 g$ and specific heat $0.1 k c a l / k g^{\circ} C$ . It contains $250 g m$ of liquid at $30^{\circ} C$ having specific heat of $0.4 k c a l / k g^{\circ} C$ . If we drop a piece of ice of mass $10 g$ at $0^{\circ} C$ . What will be the temperature of the mixture?

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Solution

Solution
Let the final temperature attained be T. then,
heat given by calorimeter + heat given by liquid
= heat absorbed by ice
$\Rightarrow M_{C} S_{C} (30 - T) + M_{L} S_{L} (30 - T)$
$= M i [L_{f u s i o n} + S_{W} (T - 0)]$
$\Rightarrow (100) (0.1) (30 - T) + (250) (0.4) (30 - T)$
$= 10 [80 + 1 (T - 0)]$
$\Rightarrow (110) (30 - T) = 10 [80 + T]$
$\Rightarrow 330 - 11 T = 80 + T$
$\Rightarrow T = \frac{(330 - 80)}{12} = {20.83}^{\circ} C$

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Q. A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg

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Q. Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3°C and B by 5°C. Which object has more specific heat? And by what factor?
b. Liquid ammonia is used in ice factory for making ice from water. If water at 20°C is to be converted into 2 kg ice at 0°C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia= 341 cal/g)
c. A thermally insulated pot has 150 g ice at temperature 0°C. How much steam of 100°C has to be mixed to it, so that water of temperature 50°C will be obtained? (Given : latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
d. A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg °C. It contains 250 gm of liquid at 30°C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0°C, What will be the temperature of the mixture?

Q. Water of mass

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t = 10^{o} C

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m_{3} = 2 k g

and temperature

- 11^{o} C

is dropped into the calorimeter. Neglecting any heat loss, the final temperature of the system is [specific heat of copper

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, specific heat of water

= 1 K c a l / k g^{o} C

, specific heat of ice

= 0.5 K c a l / k g^{o} C

, latent heat of fusion of ice

= 78.7 K c a l / k g

]

A copper calorimeter of mass 100g contains 150g of water and 10g of ice in thermal equilibrium at atmospheric pressure. A piece of lead of mass 100g at 200°C is dropped into calorimeter . Find the final temperature (assume no heat is lost to the surrounding) and the amount of ice left. (Take L=335 kJ/kg), if required.

A piece of ice of mass 40 g is dropped into 200 g of water at 50 $^{\circ} C$ . Calculate the final temperature of the water after all the ice has melted. (Specific heat capacity of water = 4200 J/kg $^{\circ} C$ , specific latent heat of fusion of ice $= 336 \times 10^{3}$ J/kg)

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A calorimeter has mass 100g and specific heat 0.1 kcal/kg∘C. It contains 250gm of liquid at 30∘C having specific heat of 0.4 kcal/kg∘C. If we drop a piece of ice of mass 10g at 0∘C. What will be the temperature of the mixture?

SolutionLet the final temperature attained be T. then,heat given by calorimeter + heat given by liquid= heat absorbed by ice⇒MCSC(30−T)+MLSL(30−T)=Mi[Lfusion+SW(T−0)]⇒(100)(0.1)(30−T)+(250)(0.4)(30−T)=10[80+1(T−0)]⇒(110)(30−T)=10[80+T]⇒330−11T=80+T⇒T=(330−80)12=20.83∘C

A calorimeter has mass $100 g$ and specific heat $0.1 k c a l / k g^{\circ} C$ . It contains $250 g m$ of liquid at $30^{\circ} C$ having specific heat of $0.4 k c a l / k g^{\circ} C$ . If we drop a piece of ice of mass $10 g$ at $0^{\circ} C$ . What will be the temperature of the mixture?