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lys-0071 [83]
2 years ago
11

The same amount of heat transferred into identical masses of different substances produces different temperature changes. Calcul

ate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of steel, originally at 20.00; c = 0.108 kcal/kg 0C .
29.3 0 C


26.9 0 C


32.3 0 C


27.4 0 C
Chemistry
1 answer:
vivado [14]2 years ago
8 0

The assumption is Originally at 20°C

Now

  • m=1
  • Q=1
  • c=0.108

We need ∆T

  • Q=mc∆T
  • 1=1(0.108)∆T
  • 0.108∆T=1
  • ∆T=1/0.108=9.26°C

Now

  • ∆T=T_f-20
  • 9.26=T_f-20
  • T_f=29.26°C

Option A

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

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            \Rightarrow T_{2}=260K

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So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
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