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Artyom0805 [142]
2 years ago
15

if the same amount of heat is added to 50.0 g samples of each of the metals, which are all at the same temperature, which metal

will reach the highest temperature?
Chemistry
1 answer:
Salsk061 [2.6K]2 years ago
6 0

The metal which will reach the highest temperature is the metal with the lowest specific heat capacity.

<h3>What is the amount of heat added to each metal?</h3>

The amount of heat Q = mcΔT where

  • m = mass of metal
  • c = specific heat capacity of mateal and
  • ΔT = temperature change

<h3>Temperature change of the metal</h3>

Making ΔT subject of the formula, we have

ΔT = Q/mc

Given that Q and m are the same for each metal,

ΔT ∝ 1/c

We see that the temperature change is inversely proportional to the specific heat capacity.

Since the metals are at the same temperature, the metal which will reach the highest temperature is the metal with the lowest specific heat capacity.

So, the metal which will reach the highest temperature is the metal with the lowest specific heat capacity.

Learn more about temperature here:

brainly.com/question/16559442

#SPJ12

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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

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\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

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\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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