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3 years ago

How much heat is required to raise the temperature of 45.5 g of water from its melting point to its boiling point?

1 answer:

3 0

First, melt the ice, H2O(s):

45.5gH2O x 333J/1gH2O = 15,152 J 

Next, raise the temperature of the water, H2O(l): 

45.5gH2O x 100degC x 4.184J/1gH2O-1degC = 19,037 J 

Finally, addem all up: 

34,200 J = 34.2 kJ to three significant figures

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A tall tube is evacuated and its stopcock is closed. The open end of the tube is immersed in a container of water (density 10^3

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Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

$p=\rho gh$

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

$h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m$

∴ The water will rise by 10.19 m.

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wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

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Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

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Answer:

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