calculate the change in entropy for 024kg of each of the fol

calculate the change in entropy for 0.24kg of each of the following as it is heated from 295k to 296k (use Table 8.1)(a)Water (b) gold (c) marble (d) wood

Solution

(a) mass m = 0.24 kg

Tempratures T = 295 K

                 T \' = 296 K

Change in entropy dS = mC ln(T \' / T )

                                = (0.24 kg) (4.19 kJ /kg K ) ln(296/295)

                                = 3.403 x10 ^-3 kJ / K

                                = 3.403 J /K

(b).mass m = 0.24 kg

Tempratures T = 295 K

                 T \' = 296 K

Change in entropy dS = mC ln(T \' / T )

                                = (0.24 kg) (0.030 kCal /kg K ) ln(296/295)

                                = 2.436 x10 ^-5 kCal / K

                                = 2.436x10 ^-2 Cal /K

(c).mass m = 0.24 kg

Tempratures T = 295 K

                 T \' = 296 K

Change in entropy dS = mC ln(T \' / T )

                                = (0.24 kg) (0.21 kCal /kg K ) ln(296/295)

                                = 1.7055 x10 ^-4 kCal / K

                                = 0.17055 cal /K

(d).mass m = 0.24 kg

Tempratures T = 295 K

                 T \' = 296 K

Change in entropy dS = mC ln(T \' / T )

                                = (0.24 kg) (1.8 kJ /kg K ) ln(296/295)

                                = 1.46 x10 ^-3 kJ / K

                                = 1.46 J /K