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

1.586 J/g°C

Step-by-step 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.