Question

A piece of silver is heated and added to a styrofoam cup calorimeter containing 26.0 mL (the density of water is 1 g/mL) of water at 19 oC. The water reaches a maximum temperature of 44 oC after the metal was added. If the specific heat of the silver is 0.240 J/g oC and it started at 93.8 oC

Answer 1

m(Ag) = 227.5g

Step-by-step explanation:

...what is the mass of the piece of silver?

The heat given for the piece of silver is the same heat that is absorbed for the water, that is:

m(Ag)×C(Ag)×ΔT(Ag) = m(Water)×C(Water)×ΔT(Water)

m is mass of water in solution: 26.0mL = 26g of water

And m(Ag) is the mass of Ag, our incognite.

C is specific heat (0.240J/g°C for Ag), and 4.184J/g°C for water

ΔT is change in temperature (93.8°C - 44°C = 49.8°C) for Ag ad (44°C - 19°C = 25°C)

Replacing:

m(Ag)×C(Ag)×ΔT(Ag) = m(Water)×C(Water)×ΔT(Water)

m(Ag)×0.240J/g°C×49.8°C = 26g×4.184J/g°C×25°C

m(Ag)* 11.952J/g = 2719.6J

m(Ag) = 227.5g