550 g of water at 105°c is poured into an 855 g aluminum container with an initial temperature of 11°c. the specific heat of aluminum is 900 j/(kg k). how much heat flows from the water to the aluminum, assuming no heat is exchanged with the surroundings?

54.22 kJ

Explanation:

In this case, we need to calculate the heat. The expression to use is:

Q = m * C * ΔT

Now, with the specific heat of water (4186 J/ kg K), we can calculate the temperature in which this occurs.

So:

Q = 0.550 * 4186 * (105 - T)

Q = 2302.3 (105 - T)

Q = 241,741.5 - 2302.3T

Now with the Aluminium:

Q = 0.855 * 900 * (T - 11)

Q = 769.5T - 8464.5

Now, with both equations, we solve for the final temperature:

769.5T - 8464.5 = 241,741.5 - 2302.3T

(2302.3 + 769.5)T = 241,741.5 + 8464.5

3071.8T = 250,206

T = 81.45 K

This is the temperature which the change occurs. Now, let's determine the amount of heat from water to Al:

Q = 241,741.5 - 2302.3(81.45)

Q = 54,219.17 J or simply 54.22 kJ.

a=d/t

a="triangle"v/t

explanation:

a = acceleration