In an experiment, 1.50 kg of concrete at 52.0$\degree$C is quickly transferred to 805 g of water at 19.0$\degree$C.The final temperature comes to 28.3$\degree$C.What is the approximate specific heat capacity of concrete? [c_p(water) =4.18 J/(g .$\degree$C)]

Question

Quizplus · Accepted Answer

The specific heat capacity of concrete can be calculated using the principle of conservation of energy, where the heat lost by the concrete is equal to the heat gained by the water. Using the formula $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature, we can set up the equation for the concrete and water and solve for the specific heat capacity of concrete. Given the final temperature and the initial temperatures of both the concrete and the water, along with the specific heat capacity of water (4.18 J/(g°C)), we can find that the specific heat capacity of concrete that fits the scenario is approximately 0.880 J/(g°C).

In an Experiment, 1 °\degree° C Is Quickly Transferred to 805 G of Water at C

In an Experiment, 1 $\degree$ C Is Quickly Transferred to 805 G of Water at C