A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

Question

OverLord2011

2 months ago

8

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

ature of the water is 24.5Â°C. Which of the following might be the metal that was used?
A) Aluminum (c=0.89J/gÂ°C)
B) Iron (c=0.45J/gÂ°C)
C) Copper (0.20J/gÂ°C)
D) Lead (c=0.14J/gÂ°C)

Chemistry

1 answer:

lions [2.9K] · Answer 1 · 2026-04-07T10:24:25+03:00

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

It is given that the heat gained by water equals the heat lost by the metal

Thus,

With the negative sign indicating heat loss

Alternatively, $m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For the metal: $m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$ Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal =?

Hence,

This value aligns with iron. Therefore, the answer is B.