A 151.6-g sample of a metal at 75.3°C is added to 151.6 g at 15.6°C. The temperature of the water rises to 18.9°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/°C·g.

4.60 J/°C g

Explanation:

This a simple calorimetry excersise. If all the heat, which is lost by the metal, is gained by the water, we assume that

Q from water = Q from metal

Q = m . C . ΔT

where C is heat capacity and ΔT, the differences between the temperatures. Let's determine the heat gained by water.

Q = 151.6 g . 4.18 J /°C g . (18.9°C - 15.6°C)

Q (+) = 2091 Joules

As this heat, was gained by the water, this heat was lost by the metal (-)

- 2091 Joules  = 151.6 g  . C  . (Final T° - 75.3°C)

We do not know at what T° was the meta, by the end, but all the heat was gained from the water, as water was increased by 3°C, metal decreases -3°C

- 2091 Joules  = 151.6 g  . C  . - 3°C

-2091 J / -3°C . 151.6g = 4.60 J/°C g

definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.