Ice cube trays hold about 350 grams of water. If the water comes out of the faucet at 37 degrees celcius, how much evergy do I have to remove to freeze the ice cubes?

-171.79 KJ

Explanation:

We must keep in mind that water freezes at 0 degrees centigrade

Recall that heat lost during freezing can be obtained from;

H = mc(θ2 - θ1)

m = mass of the water = 350 g/1000 = 0.35 Kg

c = specific heat capacity of water = 4,200 J/kg°C

θ2 = final temperature = 0°C

θ1 = initial temperature = 37°C

H= 0.35 * 4,200 * (0 - 37)

H = -54390 J = -54.39 KJ

Energy given out when 1 mole of water freezes at its freezing point = -6.04 kJ

For 350 g of water; 350 g/18 g/mol *  -6.04 kJ  = -117.4 kJ

Hence total energy removed = (-54.39 KJ) + (-117.4 kJ) = -171.79 KJ

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.