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Vladimir79 [104]
3 years ago
6

101. Du

Biology
1 answer:
n200080 [17]3 years ago
3 0

Answer:

0.972 J/g^{\circ}C

Explanation:

When the sample of hot metal is placed inside the water, the metal gives off heat to the water: the temperature of the metal decreases while the temperature of the water increases, until the two reach an equilibrium temperature.

The heat given off by the metal is equal (in magnitude) to the heat absorbed by the water, so we can write:

Q_m=Q_w\\m_m C_m (T_m-T_e) = m_w C_w (T_e-T_w)

where:

m_m=26.0 g is the mass of the metal

C_m is the specific heat of the metal

T_m=82.25^{\circ}C is the initial temperature of the metal

T_e=28.34^{\circ}C is the equilibrium temperature

m_w=75.0 g is the mass of the water

C_w=4.186 J/g^{\circ}C is the specific heat of water

T_w=24.00^{\circ}C is the initial temperature of the water

And solving for C_m, we find the specific heat of the metal:

C_m=\frac{m_w C_w (T_e-T_w)}{m_m(T_m-T_e)}=\frac{(75.0)(4.186)(28.34-24.00)}{(26.0)(82.25-28.34)}=0.972 J/g^{\circ}C

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