Question #100122
1. How long does it take for a 12.62 g sample of ammonia (NH3) to cool from 209 K to 367 K if heated at a constant rate of 6.0 kJ/min? The melting point of ammonia is 195.42 K and the boiling point is 239.82 K. The ΔHfus = 5.66 kJ/mol and the ΔHvap = 23.33 kJ/mol. The heat capacity for liquid ammonia is 80.80 J/(mol K) and the heat capacity for gaseous ammonia is 35.06 J/(mol K).
2. Calculate the heat released when 0.2 moles of ethanol (C2H6O) are heated from 100 K to 400 K. The boiling point of ethanol is 351 K and the melting point is 159 K. The heat capacity of solid, liquid and gaseous ethanol are 111.46 J/(mol•K), 112.4 J/(mol•K), 87.53 J/(mol•K). The enthalpies for ethanol are: ΔHfus = 4.9 kJ/mol and the ΔHvap = 38.56 kJ/mol.
3. How much heat is absorbed by a copper penny with a mass of 3.10 g whose temperature is raised from -8.0 °C to 37 °C? The specific heat of copper is 0.385 J/(g•K).
1
Expert's answer
2019-12-09T08:04:28-0500

1.Ammonia

n(NH3)=mM=12.6217.03=0.741moln(NH_3) = \frac{m}{M} =\frac{12.62}{17.03} = 0.741 mol

1) 209K239.82K209K\rightarrow 239.82 K

Q1=cliqiudnΔT=80.80×0.741×(239.82209)=1845JQ_1 = c_{liqiud}n\Delta T = 80.80\times0.741\times(239.82-209)=1845 J

2) vaporisation at T= 239.82 K

Q2=n×ΔHvap=0.741×23330=17288JQ_2 = n\times \Delta H_{vap} = 0.741\times 23330 = 17288 J

3) 239.82K367K239.82K \rightarrow 367 K

Q3=cgasnΔT=35.06×0.741×(367239.82)=3304JQ_3 = c_{gas}n\Delta T = 35.06\times 0.741\times (367-239.82) = 3304 J


Qtotal=Q1+Q2+Q3=1845+17288+3304=22437JQ_{total} = Q_1 +Q_2+Q_3 = 1845+17288+3304 = 22437 J


As v=6kJmin=6000Jmin=6000J60s=100Jsv = 6\frac{kJ}{min} = 6000\frac{J}{min} = \frac{6000J}{60 s} = 100\frac{J}{s} , then

time =22437100=224s=3min44s=\frac{22437}{100} = 224 s = 3 min 44 s


2.Ethanol

1) 100K159K100K \rightarrow 159K

Q1=csolid×n×ΔT=111.46×0.2×(159100)=1315JQ_1 = c_{solid}\times n \times \Delta T = 111.46\times0.2\times(159-100) = 1315J

2) melting at T= 159 K

Q2=n×ΔHfus=0.2×4900=980JQ_2 = n\times \Delta H_{fus} = 0.2\times 4900 = 980 J

3) 159K351K159K \rightarrow 351 K

Q3=cliqiud×n×ΔT=112.4×0.2×(351159)=4316JQ_3 = c_{liqiud}\times n\times \Delta T = 112.4\times 0.2 \times (351-159) = 4316 J

4) vaporisation at T=351 K

Q4=n×ΔHvap=0.2×38560=7712JQ_4 = n\times \Delta H_{vap} = 0.2 \times 38560 = 7712 J

5) 351K400K351 K \rightarrow 400K

Q5=cgas×n×ΔT=87.53×0.2×(400351)=858JQ_5 = c_{gas}\times n \times \Delta T = 87.53 \times 0.2 \times (400-351) = 858 J


Qtotal=Q1+Q2+Q3+Q4+Q5=1315+980+4316+7712+858=15181J=15.18kJQ_{total} = Q_1 + Q_2+ Q_3 + Q_4 + Q_5 = 1315 + 980+4316+7712+858 = 15181 J = 15.18 kJ


3.Copper penny

Q=cmΔT=0.385×3.10(37(8))=5371JQ=cm\Delta T = 0.385 \times 3.10 (37-(-8)) = 5371J


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