Question #108288
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
1
Expert's answer
2020-04-08T10:38:53-0400

Solution^

q=mcΔTq= m c ΔT


The heat that a copper cube loses

q1=(100/1000kg)×(400J/kg°C)×[(300Tf)°C]q_1 = (100/1000 kg) × (400 J/kg°C) × [(300 - T_f) °C]

q1=(1200040Tf)Jq_1 = (12000 - 40T_f) J


Heat received by a glass with water

q2=[(200+800)J/°C]×[(Tf25)°C]q_2= [(200 + 800) J/°C] × [(T_f - 25) °C]

q2=(1000Tf25000)Jq_2 = (1000T_f - 25000) J


q1=q2q_1 = q_2

1200040Tf=1000Tf2500012000 - 40T_f = 1000T_f - 25000

1040Tf=370001040T_f = 37000

Tf=35.6T_f = 35.6


Answer: 35.6


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