A 5 gram piece of aluminum foil at 100 degrees C is dropped into a 25 gram container of water at 20 degrees C. What is the final temperature of the aluminum.

23.30°C

Explanation:

Heat lost by Al = Heat gain by water

5× ( 100- x ) × ( 0.900) = 25× ( x-20)× ( 4.186)

(500-5x)0.900 = (25x-500)4.186

450-4.5x = 104.65x- 2093

Collecting like terms

450+2093 = 104.65x+4.5x

2543 = 109.15x

X = 2543/109.15

X = 23.298

X = 23.30°C

Final temperature of the solution is 23.3°C

Explanation:

Mass of aluminium(M1) = 5g

Initial temperature of aluminium(T1) = 100°C

Mass of water (M2) = 25°C

Initial temperature of water(T2) = 20°C

Final temperature of the solution (T3) = ?

Specific heat capacity of aluminium = 0.9J/g°C

Specific heat capacity of water = 4.184J/g°C

Heat loss by Aluminium = Heat gained by water

Q = MC∇T

Q = heat energy

M = Mass

C = specific heat capacity

∇T = change in temperature

M1C1(T1 - T3) = M2C2(T3 - T2)

5 * 0.9 *(100 - T3) = 25 * 4.184 (T3 - 20)

4.5 * (100 - T3) = 104.6 (T3 - 20)

450 - 4.5T3 = 104.6T3 - 2092

Collect liketerms

450 + 2092 = 104.6T3 + 4.5T3

2542 = 109.1T3

T3 = 2542 / 109.1

T3 = 23.299°C = 23.3°C

Final temperature of the mixture is 23.3°C

i'm assuming that the correct answer is indeed the first one/ "a"..