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maria [59]
3 years ago
6

Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject

to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)
Chemistry
1 answer:
NeTakaya3 years ago
8 0

Answer:

Rock A will have far more chemical weathering than Rock B due to the rise in area effect

Explanation:

Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.

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A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th
Bess [88]

Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

           $ = 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

           = -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

           $ = 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

           = 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water} $

        = -9.09 + 10.76

         = 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

            = 63.94 -13.22

            = 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

            = 216.39 - 9677.95

            = -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} = $

$X_{destroyed} = X_{initial} - X_{final}$

                = 50.72 + 9461.56

                = 9512.22 kJ

6 0
3 years ago
Igneous rocks with an andesitic composition ________.
slega [8]
Are found along volcanic island arcs
6 0
3 years ago
What is a common factor associated with most severe storms?
Volgvan
A common factor is low pressure system. 
4 0
3 years ago
Read 2 more answers
1kg of hydrogen is burned in oxygen how much energy is released
Simora [160]

Answer:

121 million joules energy is released

Explain :

As there are 500 moles of hydrogen gas in a kilogram, this means that burning a kilogram of hydrogen gas releases 500 times as much energy or 121 million joules .

6 0
2 years ago
Describe Chemical and physical changes of ice water placed in the refrigerator
Talja [164]
So in a chemical change, the product would loses its properties, would form new products and would not be easily reversed.


In a physical change, the products WILL NOT lose any of their properties, WILL NOT form any new products and WILL be easily reversed.

Talking about Water turning to ice I would say it’s all about physical change and not chemical change.
But I would suggest you to either make a research about it or ask your teacher :))
6 0
3 years ago
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