Question

A 23.5 g metal rod, initially at 22.7 °c, is submerged into an unknown mass of water at 63.2 °c, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 °c. The specific heat capacity of the metal is 0.544 j/g·°c and water is 4.18 j/g·°c. What is the mass, in grams, of the water?

Answer 1

The mass of the water in the insulated container, calculated using the heat transfer formula and the principle of conservation of energy, is approximately 48.9 grams.

Step-by-step explanation:

To find the mass of water in the scenario described, we can use the principle of conservation of energy, which states that the heat lost by the metal rod will be equal to the heat gained by the water. The formula for heat transfer is Q = mcΔT, where 'm' is mass, 'c' is specific heat capacity, and ΔT is the change in temperature.

Let the mass of water be 'm'. For the metal rod, we have (23.5 g)(0.544 J/g°C)(59.5°C - 22.7°C) and for the water, we have (m)(4.18 J/g°C)(59.5°C - 63.2°C). Setting the heat gained by the water equal to the heat lost by the metal gives us the following equation:

23.5 g × 0.544 J/g°C × (59.5°C - 22.7°C) = m × 4.18 J/g°C × (59.5°C - 63.2°C)

After solving for 'm', we find the mass of the water to be approximately 48.9 grams.