Answer;
The temperature change for the second pan will be lower compared to the temperature change of the first pan
Explanation;
-The quantity of heat is given by multiplying mass by specific heat and by temperature change.
That is; Q = mcΔT
This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.
-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
![t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]](https://tex.z-dn.net/?f=t%3D%20-%5Cfrac%7B1%7D%7BK_d%7Dln%5B1-%5Cfrac%7BK_d%7D%7BV_m_0%7D%28k_mln%5Cfrac%7Bs_0%7D%7Bs%7D%2B%28s_0-s%29%29%5D)
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
More force needs to be applied
Answer:
1.33
Explanation:
speed of light in vacuum, c = 3 x 10^8 m/s
speed of light in medium, v = 2.26 x 10^8 m/s
The refractive index of the medium is given by
μ = speed of light in vacuum / speed of light in medium
μ = (3 x 10^8) / (2.26 x 10^8)
μ = 1.33
Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,


P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi