An insulated copper beaker of mass M contains 300 grams of water at 760 You add 120 grams of Ice at -200C. The final stable temperature is 33.70C. What is the value of M? (Sp. Heat of copper = 386 J/kg.0C, Sp. Heat of water = 4186 J/kg.0C, Sp. Heat of ice = 2110 J/kg.0C, Latent heat of fusion of ice = 334000 J/kg.)a) 0.50 kgb) 0.61 kgc) 0.65 kg d) none of these.

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An insulated copper beaker of mass M contains 300 grams of water at 76⁰ You add 120 grams of Ice at -20⁰C. The final stable temperature is 33.7⁰C. What is the value of M? (Sp. Heat of copper = 386 J/kg.⁰C, Sp. Heat of water = 4186 J/kg.⁰C, Sp. Heat of ice = 2110 J/kg.⁰C, Latent heat of fusion of ice = 334000 J/kg.)

a) 0.50 kg

b) 0.61 kg

c) 0.65 kg

d) none of these.

Expert Solution

Step 1

Principle of Calorimetry

When two bodies of unequal temperatures are kept in thermal contact, heat exchange occurs until the two bodies reach a steady common temperature. The heat lost by the hot body is equal to the heat gained by the cold body.

$H e a t L o s t b y h o t b o d y = H e a t G a i n e d b y c o l d b o d y$

If $c$ is the specific heat of a substance then the amount of heat required for $∆ T$ change of $m k g$ os that substance is given by

$Q = m c ∆ T$

Step 2

The mass of the copper beaker $= M$

Mass of the water already present in the beaker $= 300 g = 0.3 k g$

The common temperature of the water and the beaker $T_{1} = 76 ° C = 76 + 273 = 349 K$

Final stable temperature $T_{f} = 33.7 ° C = 33.7 + 273 = 306.7 K$

Specific heat of copper $c_{1} = 386 J / k g K$

Specific heat of the water $c_{2} = 4186 J / k g K$

Heat lost by the copper beaker to reach the steady temperature $= M c_{1} (T_{1} - T_{f}) = M \times 386 \times (349 - 306.7) = 16327.8 M$

Heat lost by the water to reach a steady temperature $= 0.3 \times 4186 \times (349 - 306.7) = 53120.34 J$

Total heat lost

$Q_{1} = 16327.8 M + 53120.34$

Mass of water in the ice $= 120 g = 0.12 k g$

The initial temperature of ice $T_{2} = - 20 ° C = - 20 + 273 = 253 K$

Specific heat of ice $= 2110 J / k g K$

Amount of heat gained by ice to reach its melting point $= 0.12 \times 2110 \times (273 - 253) = 5064 J$

Latent of fusion of ice $L = 334000 J / k g$

Amount of heat gained to melt the heat completely $= M L = 0.12 \times 334000 = 40080 J$

Amount of heat required by the water in ice to increase the temperature from the melting point to the steady temperature $= 0.12 \times 4186 \times (306.7 - 273) = 16928.184 J$

Total heat gained for the entire process

$Q_{2} = 5064 + 40080 + 16928.184 = 62072.184$

According to the principle of calorimetry

$Q_{1} = Q_{2}$

This gives

$16327.8 M + 53120.34 = 62072.184 \Rightarrow 16327.8 M = 62072.184 - 53120.34 = 8951.844 \Rightarrow M = \frac{8951.844}{16327.8} = 0.55 k g$

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