An iron block of mass 500grams is dropped in a styrofoam cup of negligible mass

Question

An iron block of mass 500grams is dropped in a styrofoam cup of negligible mass containing 100 grams of water at 50C. If initial temperature of iron is 400C, how much water remains in the cup?
I am confused if I used PV=nRT to solve for volume and where do i get pressure from? Can you please help with solving for mass evaporated if final temperature was calculated to be 445K.

Explanation / Answer

o need of pV= nRT. At equilibrium both iron block & water will have same temperature. Hence, the energy disspitated from iron block will heat up the water. And some of it will evaporate as well. In order for this to be true, all the system should be at 100 degrees in the final end. Let us assume that some water is still left over, that means the temperature is 100 degres and the extra heat is used for evaporating it. Hence final T = 100 Hence use that mass (iron) * (400 - 100) * specific heat of iron= mass (water) * (100 - 50)* specific heat of water + mass (evaporated ) * evaporation calorific value 500 gm = 0.5 kg. 0.5 * (400-100) * 0.46*10^3 = 0.1 * (100-50) * 4.187 *10^3 + me * 2270*10^3 0.5 * (300) * 0.46= 0.1 * (50) * 4.187 + me * 2270 => me = 0.0211740088 kg = 21.1740088 g => left over water = 50 - me = 28.8259912 = 28.825 gms But, If the final temp is 445 k (172C) ,then nothing is left over in the beaker. The heat is radiated into the air once it gets evaporated. The first method is good because, you should considfer the energy disspitated in evaporating the water too.

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