Question

A coffee cup (or constant pressure) calorimeter contains 108.0 g of water at an initial temperature of 25.0°C. 118.7 g of tin metal at a temperature of 100°C is added. The final temperature in the calorimeter is 29.2°C. What is the molar heat capacity of the tin? The molar heat capacity of water is 75.4 J / (mol•°C). Assume that the heat capacity of the coffee cup is negligible.

Answer 1

Answer:

$26.8 J/(mol^{\circ}C)$

Explanation:

The molar mass of water is

$M_m = 18.0 g/mol$

Here the mass of water is

$m=108.0g$

So the number of moles of water in the cup is:

$n=\frac{m}{M_m}=\frac{108}{18}=6 mol$

The amount of heat released by absorbed by the water in the process is:

$Q=nC \Delta T$

where

n = 6 mol is the number of moles

C = 75.4 J / (mol•°C) is the molar heat capacity of water

$\Delta T=29.2C-25.0C=4.2C$ is the change in temperature of the water

Substituting,

$Q=(6)(75.4)(4.2)=1900 J$

According to the law of conservation of energy, this is also equal to the energy released by the hot tin metal, which can be rewritten as

$Q=nC\Delta T$

where:

$n=\frac{m}{M_m}$ is the number of moles of tin, where

m = 118.7 g is the mass of tin

$M_m=118.7 g/mol$ is the molar mass of tin

So,

$n=\frac{118.7}{118.7}=1 mol$

C is the molar heat capacity of tin

$\Delta T=100C-29.2=70.8C$ is the change in temperature of the tin

Solving for C, we find the molar heat capacity of tin:

$C=\frac{Q}{n\Delta T}=\frac{1900}{(1)(70.8)}=26.8 J/(mol^{\circ}C)$