Question

A 77.4-g sample of an alloy at 93.0°C is placed into 271 g of water at 22.0°C in an insulated coffee cup. Assume that no heat is absorbed by the cup. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy in J/(g.K)? Don't include units.

Answer 1

the specific heat capacity of the alloy in J/(g.K) is
`2.154`

Step-by-step explanation:

Knowns

Final temperature
`T_(f)=31.1\°C`, mass of water
`m_(w)=271g`, specific heat capacity of water
`c_(w)=4.186J/g*\°C`, initial temperature of water
`T_(w)=22.0\°C`, mass of the alloy
`m_(A)=77.4g` and initial temperature of the alloy
`T_(A)=93.0\°C`.

Applying Formula

We are going to use the following formula
`c_(A)=\frac{m_(w)c_(w)(T_(f)-T{w})}{m_(A)(T_(A)-T_(f))} =((271g)(4.186J/g*\°C)(31.1\°C-22.0\°C))/(77.4g(93.0\°C-31.1\°C))=2.154J/g*\°C`

Finally in this case, as the specific heat capacity is measured grade per grade, conversion between Celsius and Kelvin is direct 1:1