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 = 1.2 mL = 11.2 cm ^ 3 density of water D = 1 g / cm ^ 3 mass of water m= DV = 11.2 g intiial temp t = 26 oC final temp t ' = 100 oC mass of pan M = 1.1 kg molar mass of iron M ' = 55.847 g So, No.of moles n = M / M ' = 19.69 moles Heat lost by pan = heat gain by water heat required for water = m C dt + mL                                    = [ 11.2 * 4.186 * (100-26 ) ] + [ 11.2 * 2260.44 J / g]                                    =  3469.3568+ 25316.928                                    = 28786.2848 J This is equal to heat loss by pan SO, heat loss by pan = 28786.2848 J              nC' dt = 28786.2848                     dt = 28786.2848 / [ nC ' ]                        = 28786.2848 / [ 19.69* 25.19 ]                        = 58.037 oC                                    = 28786.2848 J This is equal to heat loss by pan SO, heat loss by pan = 28786.2848 J              nC' dt = 28786.2848                     dt = 28786.2848 / [ nC ' ]                        = 28786.2848 / [ 19.69* 25.19 ]                        = 58.037 oC

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