(C) Let the final temperature at equilibrium be $T$.According to the principle of calorimetry,the heat lost by the coffee equals the heat gained by th

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(C) Let the final temperature at equilibrium be $T$.According to the principle of calorimetry,the heat lost by the coffee equals the heat gained by the milk.Heat lost by coffee = $m_c \cdot c_c \cdot (T_i - T)$Heat gained by milk = $m_m \cdot c_m \cdot (T - T_m)$Given: $m_c = 250 \ g$,$c_c = 1.00 \ cal/g^{\circ} C$,$T_i = 90^{\circ} C$,$m_m = 20 \ g$,$c_m = 1.00 \ cal/g^{\circ} C$,$T_m = 5^{\circ} C$.Equating the two:$250 \times 1.00 \times (90 - T) = 20 \times 1.00 \times (T - 5)$$250(90 - T) = 20(T - 5)$$22500 - 250T = 20T - 100$$22600 = 270T$$T = \frac{22600}{270} \approx 83.7^{\circ} C$

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

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$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established,the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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We have half a bucket $(6 \text{ litre})$ of water at $20^{\circ}C$. If we want water at $40^{\circ}C$,how much steam at $100^{\circ}C$ should be added to it?

How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)

$A$ calorimeter has a heat capacity of $100 \ J/K$ and is at $30^{\circ}C$. $100 \ g$ of water at $40^{\circ}C$ (specific heat capacity $4200 \ J/kg \cdot K$) is poured into the calorimeter. What is the final temperature of the mixture in the calorimeter in $^{\circ}C$?

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