A 400-g block of iron at 400°C is dropped into a calorimeter (of negligible heat capacity)containing 60 g of water at 30°C.How much steam is produced? The latent heat of vaporization of water is 22.6 × 10⁵ J/kg and its specific heat is 4186 J/kg ∙ K.The average specific heat of iron over this temperature range is 560 J/kg ∙ K.

Question

Quizplus · Accepted Answer

The correct answer is A.The heat lost by the iron block is equal to the heat gained by the water and the steam produced. The heat lost by the iron block is:Q_iron = m_iron * c_iron * (T_iron - T_water)= 0.4 kg * 560 J/kg * K * (400°C - 30°C)= 112 kJThe heat gained by the water is:Q_water = m_water * c_water * (T_water - T_steam)= 0.06 kg * 4186 J/kg * K * (100°C - 30°C)= 12.5 kJThe heat gained by the steam is:Q_steam = m_steam * L_v= m_steam * 22.6 * 10^5 J/kgThe total heat gained by the water and the steam is:Q_total = Q_water + Q_steam= 12.5 kJ + m_steam * 22.6 * 10^5 J/kgSetting Q_total equal to Q_iron, we get:112 kJ = 12.5 kJ + m_steam * 22.6 * 10^5 J/kgm_steam = (112 kJ - 12.5 kJ) / (22.6 * 10^5 J/kg)= 0.0044 kg= 4.4 gTherefore, the amount of steam produced is 4.4 g.

A 400-G Block of Iron at 400°C Is Dropped into a Calorimeter