Asked by Elyssa McCalister
A 2.55x10^-2kg silver ring is heated to a temperature of 84 degrees C and then placed in a calorimeter containing 5x10^-2kg of water at 24 degrees C. The calorimeter is not perfectly insulated, however, and 0.140kJ of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?
Answers
Answered by
Anonymous
heat lost = 2.55*10^-2 * Csilver *(84-T) + 0.14*10^3 Joules
heat gained = 5*10^-2 * Cwater * (T-24)
heat lost = heat gained
solve for T
heat gained = 5*10^-2 * Cwater * (T-24)
heat lost = heat gained
solve for T
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