A 0.5970-kg ice cube at-12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.5970-kg ice cube at-12.40°C is placed inside a chamber of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 6.550 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. (Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.) Number Tools × 102

Explanation / Answer

Given,

Mi = 0.597 kg ; Tc = -12.4 deg C ; Th = 365 deg C

mass of steam = mole x molar mass

Ms = 6.55 x 18 = 118 g = 0.118 kg

Let the required temp be Tf

We know that

Q(gained) = Q(lost)

Ms Cc (365 - 100) + Ms Lv + Ms Cw (100 - Tf) = Mi Ci (0 + 12.4) + Mi Lv + Mi Cw

0.118 x 2010 x 265 + 0.118 x 2.25 x 10^6 + 0.118 x 4186 x 100 - 0.118 x 4186 x Tf = 0.597 x 2030 x 12.4 + 0.597 x 3.33 x 10^5 + 0.597 x 4186 Tf

62852.7 + 265500 + 49394.8 - 493.948 Tf = 15027.684 + 198801 + 2499.042

377747.5 - 213828.684 = 2992.99 Tf

Tf = 54.77 Deg C

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