A 55.0-g piece of metal is heated in boiling water to 99.8C and then dropped into water in an insulated beaker

Question

A 55.0-g piece of metal is heated in boiling water to 99.8 C and then dropped into water in an insulated beaker. There is 225 g of water in the beaker, and its temperature before the metal was added was 21.0C. The final temperature of the metal and water is 23.1 C. Calculate the specific heat of the metal.
Note: the specific heat for water is 4.184 J/gK

Answer 1

The heat capacity of water is 4.184 J/g°K
The specific heat of water is defined as 4.186/4.186 = 1
55s(99.8 - 23.1) = 225*(23.1-21)
s = 225*2.1/(55*76.7)
The specific heat of the metal is
s = 0.112 (dimensionless)
The heat capacity of the metal is
c = 0.112*4.184 = 0.469 J/g/°K

Answer 2

First, 1mL of water = 1cc = 1gram...that is a consistent. The metal lost seventy one.5°C. The water received 8.5°C. 75g x 8.5°C x 4.184J/g/°C = 2,667J = warmth received with techniques from water. So...14.9g of metal lost 2,667 ÷ seventy one.5 = 37.3 J 14.9 ÷ 37.3 = 0.399J/g/°C = certain warmth of the metal.