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exis [7]
2 years ago
5

Given the following specific heat capacities, which material was have the largest change in temperature if 10 grams of each subs

tance absorbs 100 calories of heat?
Substance Heat Capacity - Cal / (g oC)

aluminum 0.22

copper 0.093

lead 0.0305

silver 0.056
Physics
2 answers:
Ilya [14]2 years ago
5 0

Answer:

Explanation:

Comment

You could calculate it out by assuming the same starting temperature for each substance. (You have to assume that the substances do start at the same temperature anyway).

That's like shooting 12 with 2 dice. It can be done, but aiming for a more common number is a better idea.

Same with this question.

You should just develop a rule.  The rule will look like this

The greater the heat capacity the (higher or lower) the change in temperature.

The greater the heat capacity the lower the change in temperature

That's not your question. You want to know which substance will have the greatest temperature change given their heat capacities.

Answer

lead. It has the smallest heat capacity and therefore it's temperature change will be the greatest.

ElenaW [278]2 years ago
3 0

Find the relationship

\\ \rm\Rrightarrow Q=mc\Delta T

\\ \rm\Rrightarrow 100=10c\Delta T

\\ \rm\Rrightarrow 10=c\Delta T

\\ \rm\Rrightarrow \dfrac{10}{c}=\Delta T

\\ \rm\Rrightarrow \Delta T\propto \dfrac{1}{c}

So lower the c higher ∆T

Lead has the largest change in T as it has least capacity

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Read 2 more answers
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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