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marusya05 [52]
3 years ago
13

A small hot-air balloon is filled with 1.02×106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.09

×106 l. what is the density of the heated air in the balloon?
Physics
2 answers:
liberstina [14]3 years ago
7 0
The first thing you should know for this case is that density is defined as the quotient between mass and volume.
 d = m / v
 We have two states:
 State 1:
 d1 = 1.20 g / l
 v1 = 1.02 × 106 l
 State 2:
 v2 = 1.09 × 106 l
 Since the mass remains constant, then:
 m = d1 * v1
 Then, the density in state two will be:
 d2 = m / v2
 Substituting the value of the mass we have:
 d2 = (d1 * v1) / v2
 Substituting the values:
 d2 = ((1.20) * (1.02 * 10 ^ 6)) / (1.09 * 10 ^ 6) = 1.12 g / l
 answer:
 The density of the heated air in the balloon is 1.12 g / l
Lelechka [254]3 years ago
6 0
1.12 g/L  
The total mass of the air will remain constant, but since the volume changes and density is defined as mass per volume, we can simply calculate the new density of the heated air.  
variables 
d0, d1 = density cold, density hot
 m = mass of air
 v0, v1 = volume cold, volume hot 
 d0 = m/v0 = 1.20 g/L
 d1 = m/v1 
 m/v0 = 1.20 g/L
 m = v0 * 1.20 g/L
 m/v1 = v0 * 1.20 g/L / v1
 d1 = v0 * 1.20 g/L / v1 
 d1 = 1.02x10^6 * 1.20 g/L / 1.09x10^6
 d1 = 1.02x10^6 * 1.20 g/L / 1.09x10^6
 d1 = 1.12 g/L 
 So the density of the heated air is 1.12 g/L

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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.
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Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> 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,

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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

<u>Step 2:</u> 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

<u>Step 3: </u>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}

<u>Step 4:</u> 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

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∴T_{f} = = 115.4 ⁰C

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