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4 years ago

How does specific heat affect the rate at which an object changes temperature

2 answers:

OverLord2011 [107]4 years ago

6 0

<span>How does specific heat affect the rate at which an object changes temperature </span>All else being equal,* rate of heat transfer is proportional to the square of the temp difference between source and sink. Given a step change in input, heat transfer will change abruptly; then the rate of transfer will gradually decrease as source and sink temperatures approach each other. Ideally the rate of transfer will go asymptotically to zero as the source and sink attain equal temperatures. In a real system it will go to some stable minimum as heat leaks in/out in various directions.

<span>Since a sink with lower thermal capacity (specific heat x mass) will require less heat input for a given temperature change, heat transfer in such a system will drop off more quickly than it would in one with higher thermal capacity -- the reverse of what you're proposing. At the same time, the system will reach steady-state faster because there's less total heat in the system. </span>

Once a system has stabilized in steady-state, thermal capacity is no longer an issue.

stira [4]4 years ago

5 0

If it is lower, it heats faster, if it's higher, it takes longer to heat.

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Bacteria vary somewhat in size, but a diameter of 2.9 μm is not unusual.

Sedbober [7]

Answer:

a) 6.4 x 10^-12 cm^3

b) 17 x 10^-6 mm^2

Explanation

a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3

πr

V = \frac{4}{3}*3.142* 1.15^3

∗3.142∗1.15

= 6.3715 μm^3

1 μm^3 = 10^-12 cm^3

6.3715 μm^3 = 6.3715 x 10^-12 cm^3

==> 6.4 x 10^-12 cm^3

8 0

2 years ago

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Question 5 of 32

harkovskaia [24]

The answer is D
Neon has 20.0026 element more then the other

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2 years ago

Which technology is intended to work at a distance of about 5 to 10 centimeters with transmission speeds of 250 Kbps?

Mars2501 [29]

Answer:

NFC Near Field Communication

Explanation:

The Near Field Communication is a communication protocol, for extra short distance with a maximum of 10 centimeters, but usually used in 4 to 5 cm. Its intended to be used in contactless pay systems and in transportation card. Actually has been used to transfer multimedia from cell phones and other devices. The maximum data rate is around 424 kbit/s, with mean in 250 Kbp

5 0

4 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$