A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific he
2 answers:
Answer:
c = 250.58 J/kg/
Explanation:
The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.
Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 , then;
c = Q ÷ (mΔθ)
= 1303 ÷ (1.0 × 5.2)
= 1303 ÷ 5.2
= 250.58 J/kg/
The specific heat of the object is 250.58 J/kg/.
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Explanation:
The given data is as follows.
n = 1 mol,
Q = 1500 J, R = 8.314 J/mol k
(a)
And, according to the first law of thermodynamics
And, in an isothermal process the change in internal energy of the gas is zero.
Hence, 0 = Q - W
or, W = Q
Expression for work done in an isothermal process is as follows.
W =
As W = Q, Hence expression for Q will also be given as follows.
Q =
Now,
[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]
=
= nR ln 2
=
= 5.76 J/K
Therefore, change in entropy is 5.76 J/K.
(b) As, Q =
=
= nRT ln 2
T =
=
= 260.4 K
Therefore, temperature of the gas is 260.4 K.