The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal contact and the

Question

The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal contact and the combination is thermally insulated. The final temperature of both objects is

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Ben Gia 5 months 2021-08-10T01:48:53+00:00 1 Answers 208 views 0

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    2021-08-10T01:49:57+00:00

    Answer:

    The final temperature of both objects is 400 K

    Explanation:

    The quantity of heat transferred per unit mass is given by;

    Q = cΔT

    where;

    c is the specific heat capacity

    ΔT is the change in temperature

    The heat transferred by the  object A per unit mass is given by;

    Q(A) = caΔT

    where;

    ca is the specific heat capacity of object A

    The heat transferred by the  object B per unit mass is given by;

    Q(B) = cbΔT

    where;

    cb is the specific heat capacity of object B

    The heat lost by object B is equal to heat gained by object A

    Q(A) = -Q(B)

    But heat capacity of object B is twice that of object A

    The final temperature of the two objects is given by

    T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

    But heat capacity of object B is twice that of object A

    T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

    Therefore, the final temperature of both objects is 400 K.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )