Chemistry answer the question below

Question

A 100-g metal at 110°C is placed in a constant- pressure calorimeter containing 120-g of water at 18°C. If the specific heat of the metal is 0.57 J/gC and the calorimeter is 300 grams at 18°C initially with an average specific heat of 0.30 J/gC. Calculate the final temperature of the water.

Answer 1

The heat LOST by the hot metal MUST EQUAL the heat GAINED by the water and the calorimeter. Also, the final temperature of the metal, the water and the calorimeter will all be the same as the temperature will equilibrate among the three.

First, we'll determine the heat lost by the metal:

q = mC∆T

q = heat = ?

m = mass = 100 g

C = specific heat = 0.57 J/gº

∆T = 110º - Tf where Tf = final temperature

q = (100 g)(0.57 J/gº)(110º - Tf) = 6270 - 57Tf

Next, we find the heat gained by the water and the calorimeter combined:

For water: q = mC∆T = (120 g)(4.184 J/gº)(Tf - 18º) = 502Tf - 9037

For calorimeter: q = mC∆T = (300 g)(0.30 J/gº)(Tf - 18º) = 90Tf - 1620

Combined: 502Tf - 9037 + 90Tf - 1620

Now set heat lost by metal (6270 - 57Tf) equal to heat gained by water + calorimeter and solve for Tf:

6270 - 57Tf = 502Tf - 9037 + 90Tf - 1620

16927 = 649Tf

Tf = 26.1ºC = FINAL TEMPERATURE