Answer to Question #108288 in Molecular Physics | Thermodynamics for Fai

Question

A cooper cube whose mass mc 100 g is heated to 300 oC. ) Then the cube is dropped intro glass baker whose heat capacity Cb 200 J/K, and containing water whose heat capacity CW 800 J/K at initial temperature 25 oC then the common final temperature is ( the specific heat of cooper 400 J/Kg.k

Answer 1

Solution^

$q= m c ΔT$

The heat that a copper cube loses

$q_1 = (100/1000 kg) × (400 J/kg°C) × [(300 - T_f) °C]$

$q_1 = (12000 - 40T_f) J$

Heat received by a glass with water

$q_2= [(200 + 800) J/°C] × [(T_f - 25) °C]$

$q_2 = (1000T_f - 25000) J$

$q_1 = q_2$

$12000 - 40T_f = 1000T_f - 25000$

$1040T_f = 37000$

$T_f = 35.6$

Answer: 35.6