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

Answer:280.216j/kg°C

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