The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must be removed from 0.500 kg of water to change its temperature from 24.0°C to
5.00°C?

yes

Explanation:

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

negative sign show heat releases

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

this would be about 43%