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densk [106]
3 years ago
9

An amount of energy is added to ice, raising its temperature from -10°C to -5°C. A larger amount of energy is added to the sam

e mass of water, raising its temperature from 15°C to 20°C. From these results, what can we conclude? A) More information is needed to draw any conclusion. B) The specific heat of ice is less than that of water. C) Overcoming the latent heat of fusion of ice requires an input of energy. D) The specific heat of ice is greater than that of water. E) The latent heat of fusion of ice delivers some energy to the system.
Physics
2 answers:
algol [13]3 years ago
8 0

Answer:

B) The specific heat of ice is less than that of water.

Explanation:

Specific heat of a body is the amount of heat needed to raise a unit mass of a substance by a degree rise in temperature for a unit joule of heat given to the body.

In this case, the initial heat transfer was done below the melting temperature so no state change Temperature was reached, and hence all the heat is used to raise the temperature of the ice.

Since the temperature change in both cases are the same, and the mass are the same, it can be seen that ice has a lesser heat capacity when compared to water and hence, needs a lesser amount of heat to experience the same temperature change that the water undergoes at a much higher heat.

The higher heat capacity value means more heat will be required to raise the temperature by a degree compared to that of ice.

lara [203]3 years ago
4 0

Answer:

The correct option is;

B) The specific heat of ice is less than that of water.

Explanation:

Here we have

Let the amount of energy added to the ice at -10 C to raise the temperature to -5 C be X J

Let the amount of energy added to the water at 15 C to raise the temperature to 20 C be Y J

We know that the heat required, ΔQ to raise the temperature of a substance is given by

ΔQ = m·c·Δθ

Where:

m = Mass of the substance

c = Specific heat capacity

Δθ = Temperature change

Since the mass of the ice and the water are the same, so also is the change in temperature, (-5 - (-10) = 5 and 20 - 15 = 5) we have

for m₂·c₂·Δθ₂ > m₁·c₁·Δθ₁

Where:

m₁, c₁, Δθ₁, is for the ice and m₂, c₂, Δθ₂ is for the water and

m₁ = m₂

Δθ₁ = Δθ₂

Therefore,

c₂ > c₁ = c₁ < c₂

That is the specific heat capacity of the ice is lesser than the specific heat capacity of the water.

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Answer:

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Explanation:

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\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}

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\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}

\bar{d}=4.34 cm

b) The standard deviation equations is:

\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}

If we put all this values in that equation we will get:

\sigma=0.023 cm

Then the mean diameter will be:

\bar{d}=(4.34\pm 0.023)cm

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: V=\pi R^{2}h

Then:

\rho=\frac{m}{\pi R^{2}h}

Using the values that we have, we can calculate the value of density:

\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}

We need to use propagation of error to find the error of the density.

\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}  

  • δm is the error of the mass value.
  • δd is the error of the diameter value.
  • δh is the error of the length value.

Let's find each partial derivative:

1. \frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089

2.  \frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004

3. \frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071

Therefore:

\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}

\delta\rho=0.00058

So the density is:

\rho=(0.0089\pm 0.00058) kg/cm^{3}

I hope it helps you!

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