Answer:
a. gold : 755.7°C.
b. silver: 418.7°C.
Explanation:
a. gold :
The amount of heat absorbed by gold = Q = m.c.ΔT.
where, m is the mass of Au (m = 25.0 g).
c is the specific heat capacity of Au = 0.129 J/g°C.
ΔT is the temperature difference = (final T - 27.0°C).
Q = 2.35 kJ = 2350.0 J.
∴ The amount of heat absorbed by Au = Q = m.c.ΔT = 2350.0 J.
∴ 2350.0 J = (25.0 g)(0.129 J/g°C)(final T - 27.0°C).
∴ 2350.0 J = 3.225 final T - 87.08.
∴ 3.225 final T = 2350.0 + 87.08 = 2473.
∴ final T = 2473/3.225 = 755.7°C.
b. silver:
The amount of heat absorbed by silver = Q = m.c.ΔT.
where, m is the mass of Ag (m = 25.0 g).
c is the specific heat capacity of Ag = 0.24 J/g°C.
ΔT is the temperature difference = (final T - 27.0°C).
Q = 2.35 kJ = 2350.0 J.
∴ The amount of heat absorbed by Ag = Q = m.c.ΔT = 2350.0 J.
∴ 2350.0 J = (25.0 g)(0.24 J/g°C)(final T - 27.0°C).
∴ 2350.0 J = 6.0 final T - 162.0.
∴ 6.0 final T = 2350.0 + 162.0 = 2512.
∴ final T = 2512/6.0 = 418.7°C.
The final temperature of the gold when it absorbs the heat is 755.68 ⁰C.
The final temperature of the silver when it absorbs the heat is 418.67 ⁰C.
The given parameters;
The final temperature of the gold when it absorbs 2,350 J;
Q = mcΔt
[tex]2350 = 25 \times 0.129 \times (t- 27)\\\\2350 = 3.225t - 87.075\\\\3.225t = 2437.075 \\\\t = \frac{2437.075}{3.225} \\\\t = 755.68 \ ^0c[/tex]
The final temperature of the silver when it absorbs 2,350 J;
Q = mcΔt
[tex]2350 = 25 \times 0.24 \times (t -27)\\\\2350 = 6t - 162\\\\6t = 2350 + 162\\\\6t = 2512\\\\t = \frac{2512}{6} \\\\t = \frac{2512}{6} \\\\t = 418.67 \ ^0C[/tex]
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