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.