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

the answer is either b or c

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.