chemistry homework

Question

A hot lump of 25.425.4 g of copper at an initial temperature of 62.2 °C62.2 °C is placed in 50.0 mL H2OH2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)?0.385 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

heat lost by the hot copper must equal the heat gained by the cooler water

Use q = mC∆T to solve this problem

q = heat

m = mass

C = specific heat

∆T = change in temperature

Heat lost by hot copper = q = mC∆T

q = (25.425 g)(0.385 J/gº)(62.2º - Tf) where Tf is the final temperature

Heat gained by water = q = mC∆T

q = (50 g)(4.184 J/gº)(Tf - 25.0º) assuming a density of water = 1 g/ml

Then setting these equal to each other, we have...

(25.425 g)(0.385 J/gº)(62.2º - Tf) = (50 g)(4.184 J/gº)(Tf - 25.0º)

608.9 - 9.79Tf = 209.2Tf - 5230

218.99Tf = 5839

Tf = 26.7ºC