Answer:
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Explanation:
You need the specif heat capacities of both cobalt and lead.
- Specific heat of cobalt: 0.42 J/g.ºC
- Specific heat of lead: 0.13 J/g.ºC
When the two sheets reach the thermal equilibrium their temperatures are equal.
You can use the equations for the thermal heat to find the equilibrium temperature:
Thermal heat released by the hot sheet, lead:
- Q = 16.6 kg × 0.13J/g.ºC × (63ºC - T)
Thermal heat absorbed by the cold sheet, cobalt:
- Q = 5.78 kg × 0.42J/g.ºC × (T - 11ºC)
Equal the two equations to solve for T:
- 16.6 kg × 0.13J/g.ºC × (63ºC - T) = 5.78kg × 0.42J/g.ºC × (T - 11ºC)
I remove the units for easier handling:
- 135.954 - 2.158T = 2.4276T - 26.7036
Round to 2 significant figures: 35ºC ← answer
Answer : The [α] for the solution is, -118.8
Explanation :
Enantiomeric excess : It is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.
Mathematically,
Given:
% major enantiomer = 86 %
% minor enantiomer = 14 %
Putting values in above equation, we get:
Now we have to calculate the [α] for the solution.
Thus, the [α] for the solution is, -118.8
How advanced we’ve become in science and studying elements
The specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C
How to calculate specific heat?
The specific heat capacity of a metal can be calculated using the calorimetry equation as follows:
Q = mc∆T
Where;
Q = quantity of heat absorbed
m = mass of substance
c = specific heat capacity
∆T = change in temperature
mc∆T (water) = -mc∆T (metal)
84.7 × 4.18 × 3.8 = - (22.7 × c × -72.7)
1345.375 = 1650.29c
c = 0.815J/g°C
Therefore, the specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C.
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