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A 0.0663 kg ingot of metal is heated to 241◦C

and then is dropped into a beaker containing
0.395 kg of water initially at 25◦C.
If the final equilibrium state of the mixed
system is 27.4
◦C, find the specific heat of
the metal. The specific heat of water is
4186 J/kg ·
◦ C.
Answer in units of J/kg ·
◦ C

User Foiseworth
by
6.6k points

1 Answer

1 vote

Answer:280.216j/kg°C

Step-by-step explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

User Averi Kitsch
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6.4k points