An 100-g piece of copper,initially at 295°C,is dropped into 250 g of water contained in a 300-g aluminum calorimeter;the water and calorimeter are initially at 10.0°C.What is the final temperature of the system? (Specific heats of copper and aluminum are 0.0920 and 0.215 cal/g·°C,respectively.c_w = 1.00 cal/g·°C) A) 12.8°C B) 16.5°C C) 18.1°C D) 32.1°C

Question

Quizplus · Accepted Answer

First, we need to calculate the heat gained by the water and the calorimeter (Q1).

Q1 = (mass of water + mass of calorimeter) x specific heat of aluminum x (change in temperature)
Q1 = (250 g + 300 g) x 0.215 cal/g·°C x (final temperature - 10.0°C)

Next, we need to calculate the heat lost by the copper (Q2).

Q2 = mass of copper x specific heat of copper x (change in temperature)
Q2 = 100 g x 0.0920 cal/g·°C x (295°C - final temperature)

Since energy is conserved, Q1 = Q2.

(250 g + 300 g) x 0.215 cal/g·°C x (final temperature - 10.0°C) = 100 g x 0.0920 cal/g·°C x (295°C - final temperature)

Solving for the final temperature, we get:

final temperature = 18.1°C

An 100-G Piece of Copper,initially at 295°C,is Dropped into 250