Liter Water

Posted in February 22, 2011 ¬ 11:16 amh.adminNo Comments »

A horseshoe of 0.70 kg at1200 ℃ is dropped into a bucket with a pint of water. How much water evaporates?

Full question: A blacksmith fell by 0.70 kg iron horseshoe at a temperature of 1200 ℃ in a bucket containing a pint of water to an initial temperature of 30 ℃. How much water evaporates? Suppose that the cube does not absorb heat.

The general concept is that the thermal energy in iron is going to raise the water temperature to 100 º C, the point of boiling, then go to spin a certain amount of that water into steam. Think of the horseshoe as a bank account (heat) the payment of money (thermal energy) for water. How much heat is paid to the water to increase water temperature to 100 º C? Iron heat loss = heat gained by water * * C_iron m_iron T_iron = M_water * * C_water T_water T_iron = (m_water * * T_water C_water) / (* m_iron C_iron), note that T is the temperature changes T_iron = (500 g * 4.18 J / gK 70K *) / (700 g * 0.45 J / gK) = 464.4 K Remember that this is the change in temperature of the plate for the final temperature is 1200 to 464.4 = 735.6 C. Now that the shoe is on 735.6C and water is 100 º C, the iron heat vaporization of water enters. In the end, the shoe was going to be 100 º C and it is what remains of the water. Heat lost = heat to evaporate horseshoe certain amount of water C_iron m_iron * * * m_water_vaporized T_iron = L_vaporization m_water_vaporized = (M_iron * * C_iron T_iron) / L_vaporization m_water_vaporized = (700 g * 0.450 J / gK 635.6K *) / 2260 J / g = 88.6 grams