Question

An iron horseshoe of mass 0.85 kg is taken from a furnace at 620 ◦c and dropped into 3.98 kg of water at 5.1 ◦c. if no heat is lost to the surroundings, determine the total entropy change. the specific heat of water is 4186 j/k · kg and of iron 448 j/k · kg. answer in units of j/k.

Answer 1

To determine the total entropy change when an iron horseshoe is dropped into water, we calculate the heat transferred and subsequent entropy changes for both the horseshoe and the water, and then sum these changes to find the total entropy change of the universe.

Step-by-step explanation:

To find the total entropy change when an iron horseshoe is dropped into water, we need to calculate the entropy change for both the iron and the water as the system reaches thermal equilibrium. The final temperature (Tfinal) will be somewhere between 5.1 °C and 620 °C, where no heat is lost to the surroundings, meaning heat lost by iron is equal to heat gained by water.

To solve this problem, we first calculate the heat transferred (Q) for both the iron horseshoe cooling down and the water heating up using the formula Q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. Once we have Q for iron and water, we can find the entropy change (ΔS) using the formula ΔS = Q/T for each substance, where T is temperature in Kelvin, ensuring we integrate over the range of temperatures to account for the temperature dependence of ΔS.

Finally, we will sum the entropy changes for the iron and water to find the total entropy change in the universe, which is given by the formula ΔSuniverse = ΔSiron + ΔSwater.