The final temperature of the mixture is 43.62 °C, If 200 ml of tea at 90 °C is poured into a 400 g glass cup initially at 25 °C.
To solve the question above, we apply the law of calorimetry
The law of calorimetry: which states that if there is no lost of heat the surrounding, heat lost is equal to heat gained, or it can be stated as heat absorbed by a cold body is equal to heat released by a hot body, provided there is no lost of heat to the surrounding.
The law above is expressed mathematically as
[tex]Cm(T-t) = C'm'(t-T')[/tex]............. Equation 1
Using equation 1 to solve the question,
Let: C = specific heat capacity of glass cup, m = mass of glass cup, C' = specific heat capacity of tea, m' = mass of tea, T = initial temperature of tea, T' = initial temperature of glass cup, t = final temperature of the mixture.From the question,Given: m = 400 g = 0.4 kg, C = 840 J/kg°C, m' = 200g (tea is a liquid made of water and the volume of water in ml is thesame a its mass in gram) = 0.2 kg, C' = 4186 J/kg.°C, T = 90°C, T' = 25°CSubstitute these values into equation 1 and solve for t₃[tex]0.4(840)(90-t) = 0.2(4186)(t-25)[/tex]
[tex]336(90-t) = 837.2(t-25)[/tex]
[tex]30240-336t = 837.2t-20930[/tex]
collect like terms
[tex]837.2t + 336t = 30240+20930[/tex]
[tex]1173.2t = 51170[/tex]
[tex]t = 51170/1173.2[/tex]
[tex]t = 43.62[/tex] °C
Hence, the final temperature of the mixture is 43.62 °C
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Answer:
The final temperature of the system is 41.657 °C.
Explanation:
Let consider the tea-cup system as an isolated system, that is, a system with no energy and mass interactions with the surroundings. From the perspective of the First Law of Thermodynamics, the tea releases heat, which is received by the glass cup until thermal equilibrium is reached. The following formula represents the model under assumption that process was at steady state:
[tex]\rho_{w}\cdot V\cdot c_{w}\cdot (T_{w,o}-T) + m_{g}\cdot c_{g}\cdot (T_{g,o}-T) = 0[/tex] (1)
Where:
[tex]\rho_{w}[/tex] - Density of water, in grams per mililiter.
[tex]V[/tex] - Volume of tea, in mililiters.
[tex]c_{w}[/tex] - Specific heat of water, in joules per gram-degree Celsius.
[tex]m_{g}[/tex] - Mass of the glass cup, in kilograms.
[tex]c_{g}[/tex] - Specific heat of the glass cup, in joules per gram-degree Celsius.
[tex]T_{w,o}[/tex] - Initial temperature of tea, in degrees Celsius.
[tex]T_{g,o}[/tex] - Initial temperature of the glass cup, in degrees Celsius.
[tex]T[/tex] - Final temperature of the tea-cup system, in degrees Celsius.
If we know that [tex]\rho_{w} = 1\,\frac{g}{mL}[/tex], [tex]V = 200\,mL[/tex], [tex]c_{w} = 4.186\,\frac{J}{g\cdot ^{\circ}C}[/tex], [tex]m_{g} = 400\,g[/tex], [tex]c_{g} = 0.840\,\frac{J}{g\cdot ^{\circ}C}[/tex], [tex]T_{w,o} = 90\,^{\circ}C[/tex] and [tex]T_{g,o} = 25\,^{\circ}C[/tex], then the final temperature of the tea-cup system is:
[tex]\rho_{w}\cdot V\cdot c_{w}\cdot (T_{w,o}-T) + m_{g}\cdot c_{g}\cdot (T_{g,o}-T) = 0[/tex]
[tex]\rho_{w}\cdot V\cdot c_{w}\cdot T_{w,o} +m_{g}\cdot c_{g}\cdot T_{g,o} - (\rho_{w}\cdot V\cdot c_{w}+m_{g}\cdot c_{g})\cdot T = 0[/tex]
[tex]T = \frac{\rho_{w}\cdot c_{w}\cdot T_{w,o}+m_{g}\cdot c_{g}\cdot T_{g,o}}{\rho_{w}\cdot V\cdot c_{w}+m_{g}\cdot c_{g}}[/tex]
[tex]T = \frac{\left(1\,\frac{g}{mL} \right)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (200\,mL)\cdot (90\,^{\circ}C)+(400\,g)\cdot \left(0.840\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (25\,^{\circ}C)}{\left(1\,\frac{g}{mL} \right)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (200\,mL) +(400\,g)\cdot \left(0.840\,\frac{J}{g\cdot ^{\circ}C} \right)}[/tex]
[tex]T = 41.657\,^{\circ}C[/tex]
The final temperature of the system is 41.657 °C.
what is the relation between centre of gravity and stability
Explanation:
tilting it will raise the height of its center of gravity.
write down any 5 example of conservation of momentum?
Answer:
1) Motion of air mass moving from equator northward (closer to earth axis)
2) Motion of object in orbit
3) Collision of 2 objects
4) Skater changing rotation by extension of arms
5) Motion of rocket due to velocity of expelled gas
They want a mixture of 150 grams of water at 60 degrees Celsius, using hot water at 90 degrees Celsius and cold water at 14 degrees Celsius, how many grams of water should be used per face?
Answer:
Explanation:
Here,
Final mass(m)=150g
Final Temperature(t)=60°C
Temperature of hot water(t1)=90C
Temperature of cold water(t2)=14C
Let the grams of water to be used per face be m1 grams and (150-m1) grams for hot and cold water respectively.
We know that,
Heat lost by hot body=Heat gained by cold body
or, m1 s1 Δt1=m2 s2 Δt2
or, m1 (t1-t)=(150-m1) (t-t2) [∵s1=s2, SHC of water is taken to be same]
or, m1 (90-60)=(150-m1) (60-14)
or, 30m1=6900-46m1
or, 76m1=6900
∴m1=90.79g
(150-m1)=59.21g
Hence, 90.79g and 59.21 g hot and cold water must be used respectively.
three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???
Answer:
TEJ as this is a thing you wont get
HELP ME
Which two changes to the can opener would increase its mechanical efficiency?
A shorten the paddle so that they are closer in size to the axle that they turn
B grease the rotating wheel of the opener so that there is less breaks in between moving parts
C replaced it with a straight handle the act as levers with shorter pieces of metal
D sand the cutting surface of the metal piece so that it is perfectly smooth and sharp
Answer:
I THINK IT'S D....
HOPE SO
what can be done to increase speed of rotation of a DC motor
Answer: There are multiple options:
By varying the supply voltage.
By varying the flux, and by varying the current through the field winding.
By varying the armature voltage, and by varying the armature resistance.
Helpful Link:
https://www.elprocus.com/what-are-the-best-ways-to-control-the-speed-of-dc-motor/
A motorcycle moving with velocity and 30m/s accelerate with 4m/s2 to gain velocity of 6mls. calculate its time to gain that velocity
Using first equation of kinematics
[tex]\boxed{\sf v=u+at}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{v-u}{a}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{30-6}{4}[/tex]
[tex]\\ \sf\longmapsto t=\dfrac{24}{4}[/tex]
[tex]\\ \sf\longmapsto t=6s[/tex]
một hòn đá có khối lượng 5kg bay với vận tốc 72kg/h . động lượng của hòn đá là
Answer:
Momentum = 100 Kgm/s
Explanation:
Given the following data;
Speed = 72 km/h
Mass = 5 kg
To find the momentum of the stone;
First of all, we would have to convert the value of speed to meter per seconds.
Conversion:
72 km/h to m/s = (72 * 1000)/3600 = 20 m/s
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
Momentum = mass * velocity
Substituting the values into the formula, we have;
Momentum = 5 * 20
Momentum = 100 Kgm/s
Which device is used to measure the weight of a body?
Answer:
The beam balance is a device used for the determination of the mass of a body under gravitation. Beam balance is used for high precision measurements like masses up to 250g, it consists of a pair of scale pans one at each end of a rigid beam.
we use a scale to measure the weight of our bodies
GUYS PLEASE HELP ME THIS IS FOR MY FINAL !!!
Answer:
just add co-ratios and then find the average.
Explanation:
A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the car's new velocity?
Using 3rd equation of kinematics
[tex]\boxed{\Large{\sf v^2-u^2=2as}}[/tex]
[tex]\\ \Large\sf\longmapsto v^2=u^2+2as[/tex]
[tex]\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)[/tex]
[tex]\\ \Large\sf\longmapsto v^2=576+768[/tex]
[tex]\\ \Large\sf\longmapsto v^2=1344[/tex]
[tex]\\ \Large\sf\longmapsto v=\sqrt{1344}[/tex]
[tex]\\ \Large\sf\longmapsto v=36.6m/s[/tex]
; (b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
Explanation:
(b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
A van of mass 1200kg is moving with speed of 90km/hr. It is brought in 3 seconds by applying brakes. Calculate the force applied to bring van to rest.
Answer:
Explanation:
The main equation to solve this is F = ma, where F is the force applied to the brakes with respect to its acceleration. We have the mass that we need, but we do not have the acceleration. That's the first thing we have to find. However, our velocity needs to be stated in m/s and right now it's in km/h. Converting that:
[tex]90\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s} =25\frac{m}{s}[/tex] Now we're ready to find the acceleration:
[tex]a=\frac{v_f-v_0}{t}[/tex] where the top line there translates to the final velocity minus the initial velocity.
[tex]a=\frac{0-25}{3.0}[/tex] so the acceleration is -8.3 m/s/s
We can use that now in the force equation above:
F = 1200(-8.3) and
F = -1.0 × 10⁴ N (that's 10,000 N to the correct number of sig dig's; the negative sign means that the force is being applied in the direction opposite to that which the van is currently moving)
A ball is dropped from the roof of a 25-m-tall building. What is the velocity of the object when it touches the ground? Suppose the ball is a perfect golf ball and it bounces such that the ve locity as it leaves the ground has the same magnitude but the op posite direction as the velocity with which it reached the ground How high will the ball bounce? Now suppose instead that the ball bounces back to a height of 20 m. What was the velocity with which it left the ground?
Answer:
a) [tex]h=25m[/tex]
b) [tex]v=19.8m/sec[/tex]
Explanation:
From the question we are told that:
Height [tex]h=25m[/tex]
Bounce Height [tex]h'=20m[/tex]
Generally the Kinematic equation is mathematically given by
[tex]V=\sqrt{2gh}\\\\V=\sqrt{2*9.81*25}[/tex]
[tex]V=22.1m/sec[/tex]
Therefore Height
[tex]h=\frac{V^2}{2g}\\\\h=\frac{22.1^2}{2*9.81}[/tex]
[tex]h=25m[/tex]
b)
Generally the Kinematic equation is mathematically given by
[tex]v^2=2ah[/tex]
[tex]v^2=2*9.8*20[/tex]
[tex]v=\sqrt{2*9.8*20}[/tex]
[tex]v=19.8m/sec[/tex]
how do atoms lose electrons ?
Answer:
Explanation:
By being close to an atom that will gladly take the electrons being offered.
Suppose you are talking about Be. It is in the second column. It has two outer electrons that can be given away. It will not give away one of the two remaining electrons because they are too close to the + nucleus.
Along comes a Fluorine atom. It has 7 electrons in its outer ring. The chemistry of the situation allows it to take on one of the two electrons Be is offering. It is all a matter of charges and attractions.
Another Fluorine atom will take on the remaining electron from the Be. The outer ring cannot take on more than 1 electron, but that is enough
A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off. Round your answer to the nearest hundredth rad/hr
Answer:
w = 0.01 t⁻² rad/h
Explanation:
Let's use trigonometry to find the elevation angle of the telescope, we assume that the rocket remains fixed in the telescope.
tan θ = CO / CA
the adjacent leg is CA = 16 km = 16 10³ m
Let's use kinematics to find the height of the rocket
v = y / t
y = v t
this height is equal to the opposite leg
CO = v t
we substitute
tan θ = vt / 16
tan θ = 1200/16 t
tan θ = 75 t
θ = tan⁻¹ ( 75 t )
speed is defined by
w = dθ/dt
w = [tex]\frac{1}{1 + (75 t)^2} \ 75[/tex]
as time increases we can neglect the 1 of the denominator
w = 1/75 t²
w = 0.0133 t⁻²
Using three significant figures
w = 0.01 t⁻² rad/h
for this answer the time must be given in hours
. A radio station transmitting at a frequency of 200KH, emits waves of wavelength 1.5 km.the velocity of the radio waves is
Answer:
Explanation: as,
v=f∧
v=200×10³×1.5×10³
v=2.7×10^7ms⁻¹
The velocity of the radio waves is 3 × 10⁸ ms⁻¹.
A radio station frequency = 200 KH
wavelength = 1.5 km
How the velocity of the radio waves are calculated?Radio waves in the electromagnetic spectrum has the longest wavelength and it will always be below 300GHz. The radio waves can be generated with acceleration through some charged particles. Only through the transmitter via antenna the radio waves can gets transmitted. Radio waves can be used in all the electronic devices as mobile phones, radio communication, radars and navigations.
V = f λ
Velocity, v = ( 200 × 10³ ) × ( 1.5 × 10³ )
= ( 300 × 10⁶ )
v = 3 × 10⁸ ms⁻¹.
Hence, the velocity of the radio waves is 3 × 10⁸ ms⁻¹.
Learn more about the radio waves,
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3. There is a bell at the top of a tower that is 45 m high. The bell weighs 64 kg. The
bell has joules of potential energy.
Answer:It has 8550 j energy
resistances is inversely proprtional to___of the conductor
Answer:
resistances is inversely proportional to the area of cross section of the conductor
Explanation:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area.
hope this helps you
have a nice day:)
Please help (I didn’t know what subject to put this in but it’s earth science)
