A student uses a calorimeter to determine the enthalpy of dissolving for ammonium nitrate. The student fills a calorimeter with
475 grams of water at room temperature (24 degrees Celsius). The student adds 125 grams of solid ammonium nitrate into the calorimeter and closes the lid. She allows the ammonium nitrate to completely dissolve and observes that the temperature of the water is now 7.8 degrees Celsius. She knows that the specific heat of water is 4.18 J/g °C. Using these data, calculate the enthalpy of dissolving (Hdissolving) of the ammonium nitrate in joules per gram (J/g). Work must be shown in order to earn credit..
Enthalpy change during the dissolution process = m c ΔT,
here, m = total mass = 475 + 125 = 600 g c = <span>specific heat of water = 4.18 J/g °C </span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases) <span> Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2) = -40630 kJ (Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)
Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>
By the second law of thermodynamics: Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system. Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ). Answer: B ) Heat flows from object B to object A.
