In an experiment, 0.250 kg water at 95.0$\degree$C is quickly transferred to a 0.500 kg alumina crucible at 25$\degree$C.The final temperature comes to 61.8$\degree$C.What is the approximate specific heat capacity of alumina? [c_p(water) = 4.18 J/(g .$\degree$C)]

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To find the specific heat capacity of alumina, we use the principle of conservation of energy, assuming no heat loss to the surroundings. The heat gained by the alumina crucible is equal to the heat lost by the water. The formula for heat (q) is $q = mc\Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.For water, $q_{water} = m_{water}c_{water}\Delta T_{water}$.Given: $m_{water} = 0.250 \, kg = 250 \, g$, $c_{water} = 4.18 \, J/(g\cdot°C)$, $\Delta T_{water} = 61.8°C - 95°C = -33.2°C$.For alumina, $q_{alumina} = m_{alumina}c_{alumina}\Delta T_{alumina}$.Given: $m_{alumina} = 0.500 \, kg = 500 \, g$, $\Delta T_{alumina} = 61.8°C - 25°C = 36.8°C$.Setting $q_{water} = -q_{alumina}$ (heat lost by water is gained by alumina, hence the negative sign for water's heat to indicate loss):$$250 \, g \cdot 4.18 \, J/(g\cdot°C) \cdot (-33.2°C) = -500 \, g \cdot c_{alumina} \cdot 36.8°C$$Solving for $c_{alumina}$:$$c_{alumina} = \frac{250 \cdot 4.18 \cdot (-33.2)}{500 \cdot 36.8} \approx 1.89 \, J/(g\cdot°C)$$Therefore, the approximate specific heat capacity of alumina is 1.89 J/(g·°C).

In an Experiment, 0 °\degree° C Is Quickly Transferred to a 0

In an Experiment, 0 $\degree$ C Is Quickly Transferred to a 0