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Gekata [30.6K]
3 years ago
6

How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),

ΔHvap(H2O)=40.7kJ/mol, and ΔHfus(H2O)=6.02kJ/mol.)
Physics
1 answer:
fredd [130]3 years ago
5 0

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

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QUESTION 3
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The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

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Wood on wood kinetic friction, \mu_k = 0.3

Solution;

The force of friction, F_f, acting when the crate is moving is given as

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F_f = m × g × \mu_k

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The force with which the crate moves, F = 293.7 N

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A snowboarder on a slope starts from rest and reaches a speed of 4.2 m/s after 7.3 s
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The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².

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