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प्रश्न
50 g of ice at 0°C is added to 300 g of a liquid at 30°C. What will be the final temperature of the mixture when all the ice has melted? The specific heat capacity of the liquid is 2.65 Jg-1 °C-1 while that of water is 4.2 J g-1 °C-1. Specific latent heat of fusion of ice = 336 Jg-1.
उत्तर
m1 = 50 g, t1 = 0 °C, m2 = 300 g, t2 = 30 °C
Heat energy taken by ice to melt = 50 × 336 J = 16800 J
By law of conservation of energy,
Heat energy given by liquid = Heat taken by water
300 × 2.65 × (30 - t) = 16800 + 50 × t × 4.2
⇒ 23850 - 795 t = 16800 + 210 t
⇒ - 795 t - 210 t = 16800 - 23850
⇒ - 1005 t = - 7050
⇒ t = `7050/1005 = 7.014^circ "C"`
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संबंधित प्रश्न
1 kg of ice at 0°C is heated at a constant rate and its temperature is recorded after every 30 s till steam is formed at 100℃. Draw a temperature time graph to represent the change of phase.
Fill in the following blank using suitable word:
Ice at o0C is colder than ...........at o0C.
Which contains more heat: 1 G water at 100°C or 1 g steam at 100°C? Give reason.
Which requires more heat: 1 g ice at 0°C or 1 g water at 0°C to raise its temperature to 10°C? Explain your answer.
Explain, why are big tubs of water kept in underground cellars for storing fresh fruit and vegetables in cold countries.
20 g of ice at 0°C is added to 200g of water at 20°C. Calculate the drop in temperature ignoring the heat capacity of the container. (Specific latent heat of ice = 80 cal/g)
Find the final temperature when a mass of 80g of water at 100°C is mixed with a mass of 40g of water at 25°C.
A heater supplies heat at the rate of 800 J/s. Find the time required to convert 50g of ice at -20°C into superheated steam at 140°C.
(Given: Specific heat of ice = 2.1 J/g, latent heat of ice = 340 J/gm, latent heat of steam = 2240 J/gm, specific heat of steam = 2.1 J/gm and specific heat of water = 4.2 J/gm.)
