Round off to the nearest one (units place):

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

John decided to cycle to his friend's house at a speed of 5km/h and the journey took 2

Marianna [84]

Answer:100 miles

Explanation:

5 0

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Read 2 more answers

Find the electron and hole mobilities, and the resistivity of intrinsic silicon at 300K. Is intrinsic silicon a semiconductor

tino4ka555 [31]

Answer:

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

Explanation:

Solution:

At 300K:

Let suppose mobility of electron in intrinsic semiconductor = $M_{e}$

Mobility of electron in intrinsic semiconductor is:

$M_{e}$ = 1300 $cm^{2}$ /volt.sec

Let suppose mobility of hole in intrinsic semiconductor = $M_{h}$

$M_{h}$ = 500 $cm^{2}$ /volt.sec

We know that, intrinsic silicon semiconductor has equal number of holes and electrons. So,

At 300 K

Intrinsic Carrier Concentration = 1.5 x $10^{10}$ / $cm^{3}$ = C

And,

Conductivity of intrinsic Silicon is:

σ = C x ( $M_{h}$ + $M_{e}$ ) e

e = 1.6 x $10^{-19}$ C

So, plugging in the values, we get:

σ = C x ( $M_{h}$ + $M_{e}$ ) e

σ = 1.5 x $10^{10}$ x (500 + 1300) x 1.6 x $10^{-19}$

σ = 4.32 x $10^{-6}$

So, now we can find the resistivity.

Resistivity = 1/σ

Resistivity = 1/ 4.32 x $10^{-6}$

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

7 0

3 years ago

The y-component of a projectile’s velocity is 12.1 m/s. When the projectile once again passes by the height from which it was la

Nat2105 [25]

It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it.

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What force would be required to accelerate a 1,100kg car to 0.5 m/s2

OlgaM077 [116]

Answer:

the force required to accelerate a 1,100kg car is 550N

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