All categories

3 years ago

An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°

Physics

2 answers:

MrRa [10]3 years ago

7 0

Answer:

The specific heat of aluminum is greater.

Explanation:

It lost the most heat.

zepelin [54]3 years ago

3 0

Answer:

specific heat of aluminium is more than the specific heat of iron

Explanation:

mass of both the balls, m = 100 g

heat lost by iron is 3.6 kJ and the heat lost by aluminium is 7.2 kJ.

The amount of heat required to raise the temperature of substance by 1°C of mass 1 kg is defined as the specific heat of the substance.

Here, the amount of heat required to raise the temperature of aluminium by teh same amount as that of iron and mass remains same, so it means the opecific heat of aluminium is more than the specific heat of iron.

You might be interested in

How do particles move when a transverse wave passes through a medium?

bagirrra123 [75]

<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>

4 0

3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

2 years ago

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.

Artemon [7]

The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec

8 0

3 years ago

You discover a binary star system in which one member is a 15 solar mass main-sequence star and the other star is a 10 solar mas

Sever21 [200]

Answer:

A star with 15 solar masses is too big to be a main-sequence star.

4 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: