Answer:
7.6 g
Explanation:
"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.
The heat gained by the copper, water, and ice = the heat lost by the steam
Heat gained by the copper:
q = mCΔT
q = (120 g) (0.40 J/g/K) (40°C − 0°C)
q = 1920 J
Heat gained by the water:
q = mCΔT
q = (70 g) (4.2 J/g/K) (40°C − 0°C)
q = 11760 J
Heat gained by the ice:
q = mL + mCΔT
q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)
q = 4880 J
Heat lost by the steam:
q = mL + mCΔT
q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)
q = 2452 J/g m
Plugging the values into the equation:
1920 J + 11760 J + 4880 J = 2452 J/g m
18560 J = 2452 J/g m
m = 7.6 g
I really think it could be a 23 kg
Answer:
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Answer:
an energy source (AC or DC), a conductor (wire), an electrical load (device), and at least one controller (switch).
Explanation:
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