A 83.2 g sample of metal at 90.31 oC is added to 41.82 g of water that is initially at 19.93 oC. The final temperature of both the water and the metal is 28.42 oC. The specific heat of water is 4.184 J / (goC). Calculate the specific heat of the metal.

Question

Chemistry

Damon Dodson

2 December, 23:52

A 83.2 g sample of metal at 90.31 oC is added to 41.82 g of water that is initially at 19.93 oC. The final temperature of both the water and the metal is 28.42 oC. The specific heat of water is 4.184 J / (goC). Calculate the specific heat of the metal.

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Answers (1)

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Audrina Velasquez · Answer 1

Answer: 0.2885J/g°c

Explanation:

Loss of heat of metal = Gain of heat by the water

Therefore

-Qm = + Qw

Where

Q = mΔTCp

Q = heat

M = mass

ΔT = T. f - Ti

Ti = initial temperature

T. f = final temperature

Cp = Specific heat (m is metal, w is water)

-Qm = + Qw

-[m (T. f-Ti) Cpm] = m (T. f-Ti) Cpw

Therefore

-[83.2g (28.42-90.31) Cpm] = 41.82g (28.42-

19.93) 4.194J/g°c

Cpm = 1485.54: 5149.25

Cpm = 0.2885J/g°c

Therefore, the specific heat of the metal is 0.2885J/g°c