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How much heat is needed to raise the temperature of 9g of water by 17oC?

andre [41]

Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7

6 0

2 years ago

Does anyone know how to solve series parallel and combo circuits

tamaranim1 [39]

Answer:

Your teacher is out of her/his mind, what is he thinking

Explanation:

my dear friend, i feel sorry for you, poor thing T_T

5 0

2 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

2 years ago

Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is

Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = $\frac{K\times Q}{d^2}$

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0

3 years ago

Zoe is shown two mystery boxes that are both 8,000cm3. Her teacher tells her that one mystery box is filled with rocks and the o

krek1111 [17]

Answer:

The box of rocks will have depression which can be seen without touching the box.

Explanation:

The density of rocks is very large as compared with napkins. So, the weight of the rocks will be much more greater than that of napkins.

As both boxes have same volume the heavier box will show depression on the lower surface as compared to the lighter box. So, the box of rocks will have depression which can be seen without touching the box.

5 0

2 years ago