Answer:
- 61.2°C = Initial T°
Explanation:
This is a calorimetry problem, where the heat from the water was gained by the marble.
Q = m . C . ΔT
where ΔT is final T° - initial T°, C is the specific heat and m, mass.
By the volume of water, we realize the mass (we apply density):
1 g/mL = mass / 38.7 mL
38.7 g = mass of water
Now, we need to find out the specific heat for the marble and we have Heat Capacity data
Heat Capacity = C . m
3.52 J/°C / 4 g = C → specific heat → 0.88 J/g °C
We make the equations for both heats:
m . C . ΔT from water = m . C . ΔT from the marble
38.7 g . 4.18 J/g°C ( 26.1°C - 28°C) = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
307.35 J = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
- 307.35 is a negative value, because the was has decreased the temperature, is a loss of heat, but we have to work with the positive number.
307.35 J / (4 . 0.88 °C/J) = 26.1°C - Initial T°
87.32°C = 26.1°C - Initial T°
- 61.2°C = Initial T°
