Suppose that a 17.70 gram ball of metal initially at 84.1°C is added to 18.28 grams of an unknown liquid initially at 20.1°C. Assume that the specific heat of the metal is 1.33 J/(g·°C) and the specific heat of the liquid is 3.15 J/(g·°C).

What is the final temperature of the metal in °C? Report your answer to three decimal places.

38.716° C is the  final temperature of the metal in °C.

Explanation:

Data given:

mass of metal = 17.70 grams

initial temperature = 84.1 degrees

mass of liquid = 18.28 grams

initial temperature of liquid = 20.1 degrees

specific heat of metal is = 1.33 J/g°C

specific heat of liquid is = 3.15 J/g·°C

final temperature =?

The average temperature of the water and metal in mixture = 84.1 +20.1

= 52.1 degrees

formula used is

q = m cΔT

- q metal = q water (Heat released by metal = heat absorbed by water)

-17.70 x 1.33 (T final - 84.1) =  18.28 x 3.15 x(T final - 20.1)

-23.45 T final + 1979.79 = 57.58 T - 1157.39

81.03 T = 3137.18

T final = 38. 716 degrees

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