A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of water.The water temperature rises from 15 $\degree$C to 35 $\degree$C.Given c_Cu = 0.10 cal/g*$\degree$C,and c_water = 1.00 cal/g*$\degree$C,what was the temperature of the kiln? A)500 $\degree$C B)360 $\degree$C C)720 $\degree$C D)535 $\degree$C

Question

Quizplus · Accepted Answer

Using the formula Q = mcΔT, where Q is the heat energy gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can find the heat energy gained by the water:
Qwater = (300 g) (1.00 cal/g°C) (35°C - 15°C) = 6000 cal
Since the block of copper loses the same amount of heat energy as the water gains, we can find the initial temperature of the copper block:
Qblock = (120 g) (0.10 cal/g°C) (T - 35°C) = -6000 cal
Solving for T, we get:
T = (6000 cal / (120 g * 0.10 cal/g°C)) + 35°C = 500°C
So the temperature of the kiln was 500°C. Therefore, the correct answer is (D).

A 120-G Block of Copper Is Taken from a Kiln °\degree°

A 120-G Block of Copper Is Taken from a Kiln $\degree$