0.32 g of a walnut is burned under an aluminum can filled with 58.1 ml of water. the water temperature in the can increases by 3.1 °c. how many calories did the walnut release?

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

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.