A 500.0 g piece of aluminum at 100° C is placed in 300ml of water. While in the water, the aluminum then cools to 30°C. Calculate the amount of heat lost by the aluminum. The

specific heat of water is 4.18 J/g °C and the specific heat of aluminum is 0.90 J/g °C

The amount of heat lost by the aluminum is 31,500 J

Step-by-step explanation:

Given;

mass of aluminum, m =  500 g

initial temperature of the aluminum, θ₁ = 100° C

final temperature of the aluminum, θ₂ = 30°C

specific heat capacity of water, C = 4.18 J/g °C

specific heat capacity of aluminum , C = 0.90 J/g

Heat lost by the aluminum is equal to heat gained by the water.

The amount of heat lost by the aluminum, is calculated as;

Q = MCΔθ

Q = 500 x 0.9 (100 - 30)

Q = 500 x 0.9 x 70

Q = 31,500 J

Therefore, the amount of heat lost by the aluminum is 31,500 J

step-by-step explanation:

descending order means to order the polynomials from the term with the largest exponent to the term with the smallest exponent.  

therefore, the correct answer is d.