A piece of food is burned in a calorimeter that contains 200.0g of water. If the temperature of the water rose from 65.0°C to 83.0°C, how much heat energy in joules was contained in the food? Specific heat if water is 4.184 j/g°C.

15062.4 Joules

Explanation:

The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of food = 200.0g

C = 4.184 j/g°C

Φ = (Final temperature - Initial temperature)

= 83.0°C - 65.0°C = 18°C

Then, Q = MCΦ

Q = 200.0g x 4.184 j/g°C x 18°C

Q = 15062.4 J

Thus, 15062.4 joules of heat energy was contained in the food.

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.