Answer:
Approximately 
(assuming that the melting point of ice is 
.)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.
The energy required comes in three parts:
- Energy required to raise the temperature of that
of ice from 
to (the melting point of ice.) - Energy required to turn of ice into water while temperature stayed constant.
- Energy required to raise the temperature of that newly-formed of water from to
.
The following equation gives the amount of energy 
required to raise the temperature of a sample of mass 
and specific heat capacity 
by 
:
,
where
- is the specific heat capacity of the material,
- is the mass of the sample, and
- is the change in the temperature of this sample.
For the first part of energy input, 
whereas 
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
Similarly, for the third part of energy input, 
whereas . Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
The second part of energy input requires a different equation. The energy required to melt a sample of mass and latent heat of fusion 
is:
.
Apply this equation to find the size of the second part of energy input:
.
Find the sum of these three parts of energy:
.
O.99 m long .simple pendulum time period is 2s for second formula then use formula T=2pi.rt(lenght/gravity)
Assuming it's an ideal material, the answer is A.
The speed of sound depends on the medium that the sound travels through.
Answer:
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Explanation:
Total heat content of the fat = heat content of water +heat content of the lipids
Let it be Q
the Q= (mcΔT)_lipids + (mcΔT)_water
total mass of fat M= 0.63 Kg
Q= heat supplied = 100 W in 5 minutes
ΔT= 20°C
c_lipid= 1700J/(kgoC)
c_water= 4200J/(kgoC)
then,
solving the above equation we get
m= 0.46 kg
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
