Steam at 100 0 C is passed into 20g of water at 10 0 C When water acquires a temperature of 80 0 C, the mass of water present will be: [ Take specific heat of water = 1 cal g –1 0 C –1 and latent heat of steam = 540 cal g –1 ]
Text Solution
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m(g) steam at 100° → m(g) water at 100°C + 540m ......
m(g) water at 100°C → m(g) water at 80°C + (m) (20) ......
+
m(g) steam at 100°C → m(g) water at 80° + 560m (cal) ......
20 g water at 10°C + (20) 70 → 20 g water at 80°C ......
from and
mix + 1400 cal → (20 + m) g water at 80°C + 560m (cal)
1400 = 560m
2.5 = m
Total mass of water present
= (20 + 2.5)g
= 22.5g
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