In an experiment, 30.0 g of metal was heated to 98.0 $\circ$ C and then quickly transferred to 50.0 g of water in a calorimeter. The heat capacity of the calorimeter with the water was 211 J/ $\circ$ C. The initial temperature of the calorimeter was 27.0 $\circ$ C, and the final temperature after addition of the metal was 32.5 $\circ$ C. What is the value of the specific heat capacity of the metal?

Question

Quizplus · Accepted Answer

First, calculate the heat absorbed by water in the calorimeter:
q_water = (50.0 g) x (4.184 J/g°C) x (32.5°C - 27.0°C) = 1090 J
Next, calculate the heat lost by the metal:
q_metal = -q_water = -1090 J
Now, use the equation q = mcΔT where c is the specific heat capacity of the metal:
q_metal = (30.0 g) c (32.5°C - 98.0°C) = -25350 c
Solve for c:
c = -q_metal / (30.0 g x (32.5°C - 98.0°C)) = 0.591 J/(g°C)
Therefore, the specific heat capacity of the metal is 0.591 J/(g°C). The answer is D.

In an Experiment, 30 ∘\circ∘ C and Then Quickly Transferred to 50

In an Experiment, 30 $\circ$ C and Then Quickly Transferred to 50