In an experiment, 30.0 g of metal was heated to 98.0°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/°C. The initial temperature of the calorimeter was 27.0°C, and the final temperature after addition of the metal was 32.5°C. What is the value of the specific heat capacity of the metal?

Question

Quizplus · Accepted Answer

We can use the equation Q = mcΔT to solve for the specific heat capacity (c) of the metal.

First, let's find the heat gained by the water and calorimeter:

Qwater/calorimeter = (50.0 g)(211 J/°C)(32.5°C - 27.0°C)
Qwater/calorimeter = 5,525 J

Next, let's find the heat lost by the metal:

Qmetal = -Qwater/calorimeter
Qmetal = -5,525 J

Now, let's plug everything into the equation for specific heat capacity:

Qmetal = mcΔT
-5,525 J = (30.0 g)c(98.0°C - 32.5°C)
-5,525 J = (30.0 g)c(65.5°C)
c = -5,525 J / (30.0 g)(65.5°C)
c = 0.591 J/(g · °C)

The negative sign for Qmetal indicates that the metal lost heat to the water and calorimeter. The specific heat capacity of the metal is 0.591 J/(g · °C), which is closest to choice D.

In an Experiment, 30