The formula for energy or enthalpy is:
E = m Cp (T2 – T1)
where E is energy = 63 J, m is mass = 8 g, Cp is the
specific heat, T is temperature
63 J = 8 g * Cp * (340 K – 314 K)
<span>Cp = 0.3 J / g K</span>
Answer:
B, D, E, C, A
Explanation:
We have 5 blocks with their respective masses and volumes.
Block Mass Volume
A 65.14 kg 103.38 L
B 0.64 kg 100.64 L
C 4.08 kg 104.08 L
D 3.10 kg 103.10 L
E 3.53 kg 101.00 L
The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V
ρA = 65.14 kg / 103.38 L = 0.6301 kg/L
ρB = 0.64 kg / 100.64 L = 0.0064 kg/L
ρC = 4.08 kg / 104.08 L = 0.0392 kg/L
ρD = 3.10 kg / 103.10 L = 0.0301 kg/L
ρE = 3.53 kg / 101.00 L = 0.0350 kg/L
The order from least dense to most dense is B, D, E, C, A
Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j
Answer:
Explanation:
Hello,
In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:
Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.
Best regards.
This is a straightforward question related to the surface energy of the droplet.
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
