A fixed amount of ideal gas is kept in a container of fixed volume. The absolute pressure P, in pascals, of the gas is plotted a
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Answer:
After 30 minutes, the temperature of the body is: T₁₀ = 311.60 K
After 60 minutes, the temperature of the body is: T₂₀ = 298.18 K
Explanation:
Given that:
where;
K = 0.04
M(t) = 293
Then;
Using Euler's Formula;
where;
Then;
At initial state (0);
= 360
At t₁ = 3.0 when T₀ = 360
At t₂ = 6.0 when T₂ = 0.88T₁ + 35.16
T₂ = 0.88(351.96) + 35.16
T₂ = 344.89 K
At t₃ = 9.0 when T₃ = 0.88T₂ + 35.16
T₃ = 0.88(344.89) + 35.16
T₃ =338.66 K
At t₄ = 12.0 when T₄ = 0.88T₃ + 35.16
T₄ = 0.88(338.66) + 35.16
T₄ = 333.18 K
At t₅ = 15.0 when T₅ = 0.88T₄ + 35.16
T₅ = 0.88(333.18) + 35.16
T₅ = 328.36 K
At t₆ = 18.0 when T₆ = 0.88T₅ + 35.16
T₆ = 0.88(328.36) + 35.16
T₆ = 324.12 K
At t₇ = 21.0 when T₇ = 0.88T₆ + 35.16
T₇ = 0.88(324.12) + 35.16
T₇ = 320.39 K
At t₈ = 24.0 when T₈ = 0.88T₇ + 35.16
T₈ = 0.88(320.29) + 35.16
T₈ = 317.02 K
At t₉ = 27.0 when T₉ = 0.88T₈ + 35.16
T₉ = 0.88(317.02) + 35.16
T₉ = 314.14 K
At t₁₀ = 30 when T₁₀ = 0.88T₉ + 35.16
T₁₀ = 0.88(314.14) + 35.16
T₁₀ = 311.60 K
At t₁₁ = 33.0 when T₁₁ = 0.88T₁₀ + 35.16
T₁₁ = 0.88(311.60) + 35.16
T₁₁ = 309.37 K
At t₁₂ = 36.0 when T₁₂ = 0.88T₁₁ + 35.16
T₁₂ = 0.88(309.37)+ 35.16
T₁₂ = 307.41 K
At t₁₃ = 39.0 when T₁₃ = 0.88T₁₂ + 35.16
T₁₃ = 0.88( 307.41) + 35.16
T₁₃ = 305.68 K
At t₁₄ = 42.0 when T₁₄ = 0.88T₁₃ + 35.16
T₁₄ = 0.88(305.68) + 35.16
T₁₄ = 304.16 K
At t₁₅ = 45.0 when T₁₅ = 0.88T₁₄ + 35.16
T₁₅ = 0.88(304.16) + 35.16
T₁₅ = 302.82 K
At t₁₆ = 48.0 when T₁₆ = 0.88T₁₅ + 35.16
T₁₆ = 0.88(302.82) + 35.16
T₁₆ = 301.64 K
At t₁₇ = 51.0 when T₁₇ = 0.88T₁₆ + 35.16
T₁₇ = 0.88(301.64) + 35.16
T₁₇ = 300.60 K
At t₁₈ = 54.0 when T₁₈ = 0.88T₁₇ + 35.16
T₁₈ = 0.88(300.60) + 35.16
T₁₈ = 299.69 K
At t₁₉ = 57.0 when T₁₉ = 0.88T₁₈ + 35.16
T₁₉ = 0.88(299.69) + 35.16
T₁₉ = 298.89 K
At t₂₀ = 60 when T₂₀ = 0.88T₁₉ + 35.16
T₂₀ = 0.88(298.89) + 35.16
T₂₀ = 298.18 K