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Unit I L3 (Ch 16&17) Temperature
& heat
Do Now: solve page 531#26#48#49
#26 Patm = 0 + wghh = Patm / wg = 10.1 m Patm
P=0
#48 a) less thanb) b/c balanced, Fg = FbwoodVwoodg = wVwgwood = 900 kg/m3woodVwoodg = oilVin-oilg + wVin-wgwoodVwood = oil(1-x)Vwood + w(xVwood)x = 20%
#49 Wpuck = FbpuckVpuck g = fluidVin-fluid gVpuck = (d/2)2 * hVin-fluid = (d/2)2 * hin-fluidpuck (d/2)2 * h = fluid (d/2)2 *hin-fluidhin-fluid = puck * h / fluid = 2.1 cm
Temperature scales Kelvin
T = TC + 273.15T = TC (K or C) Celsius & Fahrenheit
TF = (9/5)TC + 32 orTC = (5/9) (TF - 32)TC = (5/9) TF
Page 566 #3 37C and 310 K #6 15C
Physical Change: temperature change & phase change
Temperature changeSpecific Heats c:
c = Q/mTunit:J/(kg*C)
Q = mcTc
Phase ChangeLatent Heats
L = Q/m SI unit:J/kg Q = mL
Turning ice into steam
Find heat energy required to turn1.5 kg of -10 C ice into 110C steam, -10C ice 0C ice: Q1=mcT = 31,350 J0 C ice 0 C water: Q2 = mL = 502,500 J0 C water 100 C water: Q3 = 627,900 J100C water 100C steam: Q4=3,390,000J100C steam 110C steam: Q5 = 30,150 JTotal energy needed: Q = 4.6 *106 J
Unit I L3 HW Ch16 #5 #35
Ch17#61 #65 #89 (a)
Unit I L4 (Ch16) Physical Changes
& Review
Do Now: (chapter 15) #88 m = T/g = 3.57 kgV = (mg-T) / wg = 3.98 * 10-4 m3 block = 8.97 * 103 kg/m3 block = 820 kg/m3
Heat energy transferred from one object to another
Q = heat = energyUnit: J or cal
where 1 cal = 4.186 J
1 Cal = 1000 calorie
Thermal Expansion Linear Expansion L = L0 T: coefficient of linear expansion
SI unit: K-1 or (C)-1 Area Expansion
A 2 A T Volume Expansion
V = VT 3 V0 T
Coefficient of thermal expansion
Ch16 page 566#18 A =2*A*T
= 2*24*10-6 *(1.178/2)2(199-23)= 0.009203 cm2
A = A + A = 1.089+0.009203= 1.0985 cm2
d = 0.5914 * 2 = 1.183 cm
Temperature v.s. heat input2.0 kg of a certain substance
Use the Temperature-Heat graph, find
a) the heat needed to raise the temperature from 100 to 110 C.
b) the specific heat of the substance at this temperature range.
c) the heat needed to melt the substance completely from solid state to liquid.
d) the latent heat of fusion for this substance.
1000 J
50 J/kg* C
4000J
2000 J/kg
Ch16 page 567#38 Qlead + Qwater = 0mleadclead*(T-Tlead) +
mwatercwater*(T-Twater) = 0235 *10-3 * 128 * (T-84.2) +
177 * 10-3 * 4186 * (T-21.5)=0T = 23.7 C
Wilsons bookPage 331 #25, 26, 27, 57, 58, 59, 61, 63,
80, 82, 84Page 391#1, 9, 11, 12, 13, 33, 34, 37
UI L4 HW Ch15, 16, 17.- Study for unit test using textbook
and class note
- memorize the conversion formula between different temperature scale.
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