heat mass × ΔT 0.59 kg × (371.2 Κ−280.0 Κ) 21,100 J specific heat

UW CHEM 120 Autumn 2012 Name __________________________________________ Section _____________ WS 1: TEMPERATURE, MATTER, AND ENERGY 1. A 0.59 kg brass candlestick has an initial temperature of 280.0 K. If 21,100 J of heat is added to the candlestick to raise its temperature to 371.2 K, what is the specific heat capacity of the brass? 21,100 J
heat
specific =
=
heat
mass × ΔT 0.59 kg × (371.2 Κ−280.0 Κ)
=
21,100 J
0.59 kg × (91.2 Κ)
2 sig figs
=
1 kg
1K
1000 g
1 °C
1 kg
1K
1000 g
1 °C
3 sig figs
21,100 J
53808 g°C
= 0.392135 J/g°C
= 0.39 J/g°C
1 of 4 UW CHEM 120 Autumn 2012 2. The 5.00 fluid oz of coffee in your styrofoam coffee cup is far too hot to drink, so you decide to let it sit and cool down a little. After a while, the temperature of the coffee is 131 oF. If 27 kJ of thermal energy were lost by the coffee as it cooled to 131oF, what was the initial temperature of the coffee, in Kelvin? Assume the density and specific heat of the coffee are the same as water (1.00 g/mL and 4.18 J/goC). volume measurement
5.00 fluid oz
ΔT =
29.57 mL
1.00 g
1 fluid oz
1 mL
heat
mass × specific heat
=
=
= 147.85 g
- 27 kJ
147.85 g × 4.184 J/g°C
1000 J
1 kJ
27000 J
618.6044 J/°C
ΔT = - 43.646634 °C = - 43.646634 K
final temp = 131 °F
TC =
TF - 32
1.8
= 55 °C
TK = TC +273.15 = 328.15 K
initial temp = final temp - ΔT
initial temp = 328.15 K - (-43.646634 K)
initial temp = 371.79663 K
initial temp = 372 K
2 of 4 UW CHEM 120 Autumn 2012 3. You have 125 g of ethanol at 20. oC. You add 40.0 kcal of energy to the ethanol. What is the final temperature and phase of the ethanol after this process? energy
added = 40.0 kcal
4.18 J
1000 cal
1 cal
1 kcal
= 167,200 J
heat required to warm 125 g = mass × ΔT × specific heat
of ethanol to its boiling point
=125 g × (78.4 °C - 20 °C) × 2.440 J/g°C
=125 g × 58.4 °C × 2.440 J/g°C
= 17812 J
heat required to convert 125 g = mass × heat of vaporization
of ethanol from a liquid to a gas
=125 g × 837.0 J/g
= 104625 J
energy remaining once 125 g
ethanol has been heated to its
boiling point and converted from = 167,200 J - (17,812 J + 104,625 J)
a liquid to a gas
= 167,200 J - (122,437 J)
= 44923 J
ΔT =
heat
mass × specific heat
=
=
44923 J
125 g × 1.650 J/g°C
44923 J
206.25 J/°C
ΔT = 217.033 °C
final temp = 78.4 °C + 217.033 °C = 295.433 °C
so, the ethanol will be a gas at 300 °C
3 of 4 UW CHEM 120 Autumn 2012 Potentially helpful information. Conversion factors: Length 1 m = 1.094 yd 1 mi = 5280 ft 1 ft = 12 in 3
1 km = 10 m 1 mi = 1760 yd 1 Å = 10-‐10 m 1 m = 102 cm (centimeters) 1 in = 2.54 cm (exact) 1 m = 103 mm (millimeters) Mass 1 kg = 2.205 lb 1 lb = 453.6 g 1 amu = 1.6605 x 10-‐27 kg 1 dry oz (ounce) = 28.35 g Volume 1 L = 1000 mL = 1.06 qt 1 ft3 = 28.32 L 1 dL = 10-‐1 L 1 mL = 1 cm3 1 gal = 3.785412 L 1 fluid oz = 29.57 mL Pressure 1 atm = 760. mmHg 1 torr = 1 mmHg 2
1 atm = 14.6959 lb/in (psi) 1 atm = 101,325 Pa 1 Pa = 1 N/m2 1 bar = 100 kPa Energy 1 eV = 1.602 x 10-‐19 J 1 cal = 4.184 J (exact) 1 kcal = 1 Cal Force 1 N = 0.22481 lb Metric prefixes: tera (T) = 1012 peta (P) = 1015 pico (p) = 10-‐12 femto (f) = 10-‐15 Equations: m
ΔT = T final − Tinitial
TF = 1.8 ⋅ TC + 32
q = s ⋅ m ⋅ ΔT
d =
V
TK = TC + 273.15
Physical Data: Ethanol Water 3
3
Density, liquid = 1.000 g/cm Density, liquid = 0.789 g/cm Heat of vaporization = 837.0 J/g Heat of fusion = 107.9 J/g o
Tboil = 78.4 C o
Tfreeze = -‐114 C Specific heat, liquid = 2.440 J/g.K Specific heat, gas = 1.650 J/g.K 3
Density, solid = 0.917 g/cm Heat of vaporization = 2260 J/g Heat of fusion = 334 J/g o
Tboil = 100 C o
Tfreeze = 0 C Specific heat, solid = 2.11 J/g.K Specific heat, liquid = 4.184 J/g.K Specific heat, gas = 2.080 J/g.K 4 of 4

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