Question

A 2.00 kg piece of aluminum metal at 75.0 °C is placed in 6.00 liters (= 6.00 kg) of water at 35.0 °C. Determine the final temperature .

Answer 1

Answer:

T = 46.97 °C

Explanation:

Applying the law of conservation of energy, we get:

Heat Energy Lost by Aluminum Piece = Heat Energy Gained by Water

$m_aC_a\Delta T_a=m_wC_w\Delta T_w$

where,

$m_w$ = mass of water = (Density)(Volume) = (1000 kg/m³)(1 L)(0.001 m³/1 L)

$m_w$ = 1 kg

$m_a$ = mass of auminum piece = 2 kg

$C_w$ = specific heat capacity of water = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum = 897 J/kg.°C

$\Delta T_w$ = Change in Temperature of Water = T - 35°C

$\Delta T_a$ = Change in Temperature of Aluminum Piece = 75°C - T

T = Final Temperature = ?

Therefore,

$(2\ kg)(897\ J/kg.^oC)(75^oC - T)=(1\ kg)(4200\ J/kg.^oC)(T - 35^oC)\\\\134550\ J - (1794\ J/^oC)T = (4200\ J/^oC)T - 147000\ J\\281550\ J = (5994\ J/^oC)T\\\\T = \frac{281550\ J }{5994\ J/^oC}$

T = 46.97 °C