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

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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?