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

First, calculate the amount of heat absorbed by the water and the calorimeter:

q₁ = m₁c₁ΔT = (50.0 g)(211 J/°C)(32.5°C - 27.0°C) = 5,527.5 J

Next, calculate the amount of heat lost by the metal:

q₂ = -q₁ = -5,527.5 J

Finally, use the equation for heat (q = mcΔT) to solve for the specific heat capacity of the metal:

q₂ = mcΔT

-5,527.5 J = (30.0 g)c(98.0°C - 32.5°C)

c = 0.591 J/(g°C)

Therefore, the specific heat capacity of the metal is 0.591 J/(g°C), which corresponds to choice D.

In an Experiment, 30