A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final temperature of 51.9°C. The specific heat capacity of water is 4.186 joules per degree Celsius, and the specific heat capacity of limestone is 0.921 joules per degree Celsius.

What was the initial temperature of the limestone?

The initial temperature of the limestone could be derived by using the principle of conservation of energy and the formula for heat exchange (Q = mcΔt). Calculate the heat gained by the water, the heat lost by limestone, and equate the two to solve for the limestone's initial temperature.

Explanation:

This problem can be approached using the concept of conservation of energy, in this case, thermal energy. The heat loss or gain can be represented by the formula Q=m*c*Δt, with ‘Q’ being the amount of energy, ‘m’ the mass, ‘c’ the specific heat capacity and ‘Δt’ the temperature change. The sum of energy before and after in an isolated system (water and limestone here) stays the same. So, we can state the equation as Heat gained by water = Heat lost by limestone. The limestone's final temperature is 51.9C (given) so you can solve the equation for the limestone's initial temperature. Start by calculating the heat gained by water:

Qwater=mcΔt=75.0g*4.186j/g°C*(51.9°C-23.1°C)

Then, calculate heat lost by limestone:

Qlimestone=mcΔt=62.6g*0.921j/g°C*(t_initial - 51.9°C)

Setting Qwater = Qlimestone and solving for the unknown will give you the initial temperature of the limestone.

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