Exactly 11.2 mL of water at 26.0 degree C are added to a hotiron skillet. All of

Question

Exactly 11.2 mL of water at 26.0 degree C are added to a hotiron skillet. All of the water is converted into steam at 100degree C. The mass of the pan is 1.10 kg and the molar heatcapacity of iron is 25.19 J/(mol. degree C). What is thetemperature change of the skillet? ______ degree C Exactly 11.2 mL of water at 26.0 degree C are added to a hotiron skillet. All of the water is converted into steam at 100degree C. The mass of the pan is 1.10 kg and the molar heatcapacity of iron is 25.19 J/(mol. degree C). What is thetemperature change of the skillet? ______ degree C

Explanation / Answer

volume of water V = 11.2 mL =11.2 cm ^ 3 density of water D = 1 g / cm ^ 3 mass of water m = VD = 11.2 g Initial temp t = 26 oC final temp t ' = 100 oC temp differecne dt = 100 -26 = 74 oC Heat required to convert water from temp t to t 'is Q = m C dt                   Q = 11.2 g * 4.186 J / g oC * 74 oC                       =     3469.3568 J Mass of pan m ' = 1.1 kg = 1100 g molar mass of iron M = 56 g So, No.of moles n = m ' / M                             = 19.642 mol molar heat capacity of iron C ' = 25.19J/(mol. degreeC).                          we know from problem Q = nC' dt ' from this the temperature change of the skillet dt ' = Q/ nC '                                                                           = 7.011 oC

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.