Answer:
Specific heat of solid is 0.401J/g°C
Explanation:
In this experiment, the heat losed for the unknown solid is equal to heat gained for the water.
Using calorimeter equation:
Q = m×C×ΔT
Where Q is heat, m is mass of substance, C is specific heat and ΔT is change in temperature.
It is possible to write:
m[Solid] × C[Solid] × ΔT[Solid] = m[Water] × C[Water] × ΔT[Water]
Replacing:
77.5g × C[Solid] × (62.5°C - 26.2°C) = 93g × 4.184J/g°C × (26.2°C - 23.3°C)
77.5g × C[Solid] × (62.5°C - 26.2°C) = 93g × 4.184J/g°C × (26.2°C - 23.3°C)
2813.25g°C × C[Solid] = 1128.4J
C[Solid] = 0.401 J/g°C
Specific heat of solid is 0.401J/g°C
