Answer:
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
