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

i would say b

explanation:

but i think i would need more information than that