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

`26.8 J/(mol^(\circ)C)`

Step-by-step 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=(m)/(M_m)=(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=(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=(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=(Q)/(n\Delta T)=(1900)/((1)(70.8))=26.8 J/(mol^(\circ)C)`