PROBLEMS Answer the test questions - I
LA calorimeter contains water at 0 "C and pieces of floating ice. If the water is stirred with a metal bar for a while, which one of the following can be
observed?
A) Some of the 1ce pieces melt
B) Some of the water freezes.
C) All of the water freezes. D) Nothing changes
2. The amount of heat needed to change the temperature of a unit mass of a substance by 1 "C is
called:
A) Heat of vaporisation
B) Specific heat capacity C) Heat of fusion
D) Heating value
3. Two blocks of copper and aluminum have the same mass and temperature of 0 *C They are dropped into two different calorimeters. Each Which ONE of the statements is correct when calorimeters reach thermal equilibrium?
calorimeter has water of 100 g at 60 "C. Take specific heat capacity of calorimeters to be zero
A) The copper has a higher temperature than the
aluminum
B) The copper has a lower temperature than the
aluminum
C) The temperatures of the two calorimeters are the D The answer depends on the volumes of the metal
same
blocks
4 Which of the following is the process by which a solid changes directly to a gas?
A) Melting
B) Evaporation C) Freezing
D) Sublimation
5. When a sample of vapour condenses into a liquid,
A) it gains heat
B) it loses heat
its temperature rises D) its temperature drops
6 In the SI unit system heat is measured in
A) Newtons
B) Kilograms C) Metres
D) Joules
7. During boiling of water,
the molecules of water move faster. I II. the temperature of water increases.
III the water molecules move slowly
IV. the temperature doesn't change Which of the statements above are correct?
A) I and II
B) I, II, IV C)l and IV
D) II, III and IV
8.100 g of water at 0 C is added to 80 g of water at
90 "C. Calculate the final temperature of the mixtu
A) 20 C
B) 40 °C C) 45 °C
D) 60 C
Answer the questions 9, 10 and 11 looking at the information below.
An iron block at 200 "C is dropped into a calorimeter which contains 50 g of water. The calorimeter is tightly closed. The graph below shows the
temperature changes of water
Temperature IC
140
100
60
Time Is
9 What is the final temperature of the iron?
A) 60 °C B) 100 C
C) 140 "C D) 200 °C
10. How much energy did the water gain when it became vapour at 140 C?
A) 24 kJ
B) 27 kJ C) 28 kJ
D) 125 kJ
IL What is the mass of the iron?
A) 5 kg B) 2 kg
C) 200 g D) 500 g
Определим энергию которая необходима для плавления льда взятого при температуре плавления:
Qл = m1∙λ (1).λ = 33∙104 Дж/кг.
Qл = 16,5∙104 Дж.
Определим количество теплоты которая выделится при остывании воды от 80 0С до 0 0С:
Qв = с∙m2∙(t2 – t1) (2).с = 4200 Дж/кг∙0С.
Qв = -3,36∙104 Дж.
Количество теплоты которое выделится при остывании воды от 80 0С до 0 0С меньше чем энергия которая необходима для плавления льда взятого при температуре плавления.
Часть льда расплавится, часть останется в твердом состоянии. Энергия которая выделится при остывании воды пойдет на плавление части льда.
Определим массу льда который расплавится:QB=m⋅λ, m=QBλ.m = 0,102 кг.
1.Найдите:
А) Амплитуду колебаний заряда.
В общем виде уравнение колебаний заряда q=qm*cos(ωt). Cопоставляя получаем qm=5*10^-4 Кл.
Б) Период. ω= 10^3π. Из ω = 2π/T, T=2π/ω=2π/(10^3π)=2*10^-3 c.
В) Частоту. Из υ=1/T, υ=1/(2*10^-3) =0,5*10^3 Гц= 500 Гц.
Г) Циклическую частоту. ω= 10^3π Гц= 3140 Гц.
2. Запишите уравнения зависимости напряжения на конденсаторе от времени:
Из формулы емкости конденсатора С=q/U имеем
u(t) = q(t)/C =
(5*10^-4cos(10^3πt))/(10*10^-12) = 0,5*10^8 cos(10^3πt):
и силы тока в контуре от времени: в общем виде i(t) =q(t) '=Imcos(ωt+π/2) - ток опережает колебания напряжения на конденсаторе на π/2, Im=ω*qm; Im=10^3π*5*10^-4=1,57 A.
Значит i(t) =1,57cos(10^3πt+π/2).