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

T = 46.97 °C

Step-by-step 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 = (281550\ J )/(5994\ J/^oC)`

T = 46.97 °C