200 gm of water at 20 degree Celsius is mixed with 20 gm of steam at120 degree Celsius find mixture temperature.the process is not carried out in atmospheric pressure hence the boiling point of water is 110degree celius

Question

Kruthik · Answer

Let T be the final temperature of the mixture. Let’s calculate the amount of heat released by steam to be converted into water at T o C. First, steam at 120 o will be converted into 110 o Heat relesed during this process Q 1 =mS∂T=20g×0.46cal g -1 o C -1 × 10 o C = 92cal. Heat released during the conversion of steam in water Q 2 =mL v =20g × 540calg -1 = 10780cal Heat released during the conversion of water at 110 o C to water at T o C = mS∂T = 20g×1calg -1 o C -1 (110 – T) Now Q steam= Q 1 +Q 2 + Q 3 = (11092 + 20T) Now let’s calculate heat absorbed by water to get convert to T o C Q water = mS∂T = 200g × 1cal g -1 o C -1 × (T-20) = 200T – 4000. By principle of Calorimetry, Q steam = Q water. Hence, 11092 + 20T = 200T – 4000 So, T = 63.84 o C.

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200 gm of water at 20 degree Celsius is mixed with 20 gm of steam at120 degree Celsius find mixture temperature.the process is not carried out in atmospheric pressure hence the boiling point of water is 110degree celius

200 gm of water at 20 degree Celsius is mixed with 20 gm of steam at120 degree Celsius find mixture temperature.the process is not carried out in atmospheric pressure hence the boiling point of water is 110degree celius

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