Question

a piece of chromium metal with a mass of 24.26g is heated in boiling water to 98.3 degrees celsius and then dropped into a coffee-cup calorimeter containing 82.3g of water at 23.3 degrees celsius. when thermal equilibrium is reached, the final temperature is 25.6 degrees celsius. calculate specific heat capacity of chromium.

Answer 1

Answer on Question #55563 – Chemistry - General chemistry

Question:

A piece of chromium metal with a mass of 24.26 g is heated in boiling water to 98.3 degrees celsius and then dropped into a coffee-cup calorimeter containing 82.3 g of water at 23.3 degrees celsius. when thermal equilibrium is reached, the final temperature is 25.6 degrees celsius. Calculate specific heat capacity of chromium.

Answer:

In the considered system the energy sent is equal to the energy received.

$$Q = C \cdot m \cdot \Delta T$$ $$C_{\text{(water)}} \cdot m_{\text{(water)}} \cdot \Delta T_{\text{(water)}} = C_{\text{(chromium)}} \cdot m_{\text{(chromium)}} \cdot \Delta T_{\text{(chromium)}}$$

Where C – specific heat capacity for the material, J/(g·K);

m – mass of the material, g;

$\Delta T$ – temperature difference of the material, K.

$$C_{\text{(water)}} = 4.1813 \, \text{J/(g·K)}$$ $$C_{\text{(chromium)}} \cdot 24.26 \cdot ((98.3 + 273) - (25.6 + 273)) = 4.1813 \cdot 82.3 \cdot ((25.6 + 273) - (23.3 + 273))$$ $$C_{\text{(chromium)}} = \frac{4.1813 \cdot 82.3 \cdot ((25.6 + 273) - (23.3 + 273))}{24.26 \cdot ((98.3 + 273) - (25.6 + 273))} = 0.45 \, \frac{J}{g \cdot K}$$

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