Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
Answer:
714 nm
Explanation:
Using the equation: nλ=d<em>sin</em>θ
where
n= order of maximum
λ= wavelength
d= distance between lines on diffraction grating
θ= angle
n is 1 because the problem states the light forms 1st order bright band
λ is unknown
d= 
or 0.0000014 (meters)
sin(30)= 0.5
so
(1)λ=(0.0000014)(0.5)
=0.000000714m or 714 nm
Answer:
hydroelectric , hydrogen fuel cells , solar power , geothermal , wind power
Explanation:
all of these has little to no waste being produced and have little to no impact on the environment .
