Question

Please help!!!

A piece of metal of mass 27 g at 93° C is placed
in a calorimeter containing 59.2 g of water at
21°C. The final temperature of the mixture is 34.9 ° C. What is the specific heat capacity of the metal? Assume that there is no energy lost to the surroundings.
Answer in units of J/g. ° C

Answer 1

Answer:

1.586 J/g°C

Explanation:

So, we have the formula $q = mc\Delta t$.

Since heat released by the metal is = to the heat absorbed by the water (because they eventually become the same temperature in solution), we can say

$m_{water}C_{water}(T_{water}-T_f) = - m_{metal}C_{metal}(T_{metal} - T_f)$

Plugging in, we get:

$59.2*4.184*(21- 34.9) = - 27*C_{metal}*(93 - 34.9)$

Solving, we get $C_{metal} = 1.586$ J/g°C.