Two hundred thirty grams of water at 0°C is brought into contact with a heat reservoir at 92°C. After thermal equilibrium is reached, what is the temperature of the water (in °C)? (The specific heat of water is4,186 J/(kg · °C).)92°CAfter thermal equilibrium is reached, what is the temperature of the reservoir (in °C)?92°CHow much heat (in J) has been transferred in the process? (Enter the magnitude. Round your answer to at least four significant figures.)88575.76What is the entropy change of the water (in J/K)? (Round your answer to at least one decimal place.)279X J/KWhat is the entropy change of the reservoir (in J/K)? (Round your answer to at least one decimal place.)-2.43E-2X J/KWhat is the entropy change of the universe (in J/K)?J/K

given m=230g=0.230kg, C=4186J(kg⁰C) 1. Since reservoir has infinite capacity at same temperature. Therefore the temperature of the water reached to 92⁰. So answer is 92⁰C. 2. Temperature of reservoir remain constant. Therefore answer is 92⁰C. 3. Heat transferred in the process equal to heat gained by water =mCΔT=0.230x4186x(92-0)=88575.76J So answer is 88575.76J4. Entropy change in water is This is the answer. 5. Since the temperature of the reservoir remain constant therefore change in its entropy is .6. Now the change in entropy of the universe is S=Sw+Sr=279.62-242.59=34.06J/k This is the answer.

Answered: Two hundred thirty grams of water at…