A 100.0g sample of an unknown metal at 100.0 degrees Celcius is added to 200.0mL

Question

A 100.0g sample of an unknown metal at 100.0 degrees Celcius is added to 200.0mL of water (denisty of 1.00g/mL) at 25 degrees Celcius. The temperature of the water was measured after thermal equilibrium was reached and the final temperature of the water is 28.5 degrees Celcius. Determine the specific heat capacity of the unknown metal. A 100.0g sample of an unknown metal at 100.0 degrees Celcius is added to 200.0mL of water (denisty of 1.00g/mL) at 25 degrees Celcius. The temperature of the water was measured after thermal equilibrium was reached and the final temperature of the water is 28.5 degrees Celcius. Determine the specific heat capacity of the unknown metal.

Explanation / Answer

-Qmetal = Qwater

-mmetal * Cpmetal ( Tf- Tmetal) = Mwater * Cpwater*(Tf-Twater)

substitute

-100* Cpmetal ( 28.5- 100) = 200* 4.184*(28.5-25)

Cp metal =  200* 4.184*(28.5-25) / ( 28.5- 100) / -100

Cp metal = 0.40962J/gC

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