Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid = 
Energy released by 0.004667 moles of benzoic acid on combustion:

Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C



Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose = 
Energy released by the 0.016097 moles of calorimeter combustion:

Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'



The temperature change from the combustion of the glucose is 6.097°C.
Distance and period of time
Don't understand the text, but water always freezes at 0°C as long as its water.
Answer:
1.8 × 10² cal
Explanation:
When 0.32 g of a walnut is burned, the heat released is absorbed by water and used to raise its temperature. We can calculate this heat (Q) using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity of water
m: mass of water
ΔT: change in the temperature
Considering the density of water is 1 g/mL, 58.1 mL = 58.1 g.
Q = c × m × ΔT
Q = (1 cal/g.°C) × 58.1 g × 3.1°C
Q = 1.8 × 10² cal