A 1.001 0 2 g aluminum block at 100.0°C is placed in 1.001 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?

CA = (-mBCB∆TB) / mA∆TA

Question

Heidi Mcdonald

Physics

15 March, 07:24

A 1.001 0 2 g aluminum block at 100.0°C is placed in 1.001 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?

CA = (-mBCB∆TB) / mA∆TA

+5

Answers (1)

Know the Answer?

Brock · Answer 1

The answer to your question is Ca = 0.904 J/g°C

Explanation:

Data

Aluminum Water

mass 1 x10² g 1 x 10² g

Temperature 1 100°C 26°C

Temperature 2 10°C 10°C

Specific heat? 4.182 J/g°C

Formula

Aluminum Water

-mCa (T2 - T1) = mCw (T2 - T1)

-Solve for Ca

Ca = [mCw (T2 - T1) ] / - m (T2 - T1)

-Substitution

Ca = [1 x 10² x 4.182 x (26 - 10] / - 1 x10² (26 - 100)

-Simplification

Ca = [418.2 (16) ] / - 1 x 10² (-74)

Ca = 6691.2 / 7400

-Result

Ca = 0.904 J/g°C

A 1.00*1 0 2 g aluminum block at 100.0°C is placed in 1.00*1 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?CA = (-mBCB∆TB) / mA∆TA

A 1.001 0 2 g aluminum block at 100.0°C is placed in 1.001 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?

CA = (-mBCB∆TB) / mA∆TA