What type of friction is using chalk in the summer to draw on the ground in Copley square?

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b

Lesechka [4]

Answer:

a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,

c) U = 54.7 nJ , d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

C = Q / V

Q = C V

can also be calculated using geometry consideration

C = e or A / d

we reduce to the SI system

A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

d = 1.53 cm = 1.53 10⁻² m

we substitute

Q = eo A / d V

Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

Q = k Q₀

Q = 80 397.57

Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

C = k C₀

C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

C = 1.157 10⁻¹⁰ F

V = Vo / k

V = 275/80

V = 3.4375 V

c) The stored energy is

U = ½ C V²

U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²

U = 5.47 10⁻⁸ J

let's reduce to nJ

109 nJ = 1 J

U = 54.7 nJ

d) energy after submerging

U = ½ (kCo) (Vo / k) 2

U = ½ Co Vo2 / k

U = U₀ / k

U = 54.7 / 80 nJ

U = 0.68375 nJ

the energy change is

ΔU = U₀ -U

ΔU = 54.7 - 0.687375

6 0

3 years ago

the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating

VARVARA [1.3K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
$\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}$

4 0

3 years ago

Which property do gas particles at the same temperature share?

atroni [7]

It’s potential energy

8 0

3 years ago

Read 2 more answers

When a can is crushed did it undergo a physical change

brilliants [131]

Because even though the object got crush and misshape it still has the same identity. the identity never change

4 0

3 years ago