Question

Quando misturamos 2,0 Kg de água (calor específico sensível =1,0 cal /g° C ) a 80° C com 2,0 kg de água a 20°C, obtemos 4,0 Kg de água a:


a) 10°C

b) 20°C

c) 30°C

d) 40°C

e) 50°C

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Answer 1

Questão de equilibrio térmico :

m × c × ( Tf - Ti ) + m × c × ( Tf - Ti ) = 0

2 kg = 2000 g

2000 × 1 × ( Tf - 80 ) + 2000  × 1 × ( Tf - 20 ) = 0

2000 × ( Tf - 80 ) + 2000 × ( Tf - 20 )

2000Tf - 160.000 + 2000Tf - 40.000 = 0

2000Tf + 2000Tf - 160.000 - 40.000 = 0

4000Tf - 200.000 = 0

4000Tf = 200.000

Tf = $\frac{200.000}{4000}$

Tf = 50°C

alternativa e)

Answer 2

Explicação:

-> $\sf 2~kg=2000~g$

$\sf Q_1+Q_2=0$

$\sf m_1\cdot c_1\cdot\Delta T_1+m_2\cdot c_2\cdot\Delta T=0$

$\sf 2000\cdot1\cdot(T-80)+2000\cdot1\cdot(T-20)=0$

$\sf 2000T-160000+2000T-40000=0$

$\sf \dfrac{2000T}{1000}-\dfrac{160000}{1000}+\dfrac{2000T}{1000}-\dfrac{40000}{1000}=0$

$\sf 2T-160+2T-40=0$

$\sf 4T-200=0$

$\sf 4T=200$

$\sf T=\dfrac{200}{4}$

$\sf T=50°C$

Letra E