Question

A 0.4 kilogram sample of aluminum at 115 degrees Celsius is put into a container containing 0.5 kilograms of water at 15 degrees Celsius. Neglecting the small amount of energy absorbed by the container and knowing that the specific heat of aluminum is 900 kJ/kg*C, and the specific heat of the water is 4186 kJ/kg*C answer the following question. Compared to the heat liberated by the aluminum, the heat absorbed by the water is

Answer 1

Heat given by aluminium must be equal to heat absorbed by water

Step-by-step explanation:

When we mix two objects at different temperatures then two objects then due to temperature gradient the heat will flow from high temperature to low temperature.

The flow of heat will continue till the temperature will be same for two objects.

so here we will have

heat given by aluminium = heat absorbed by water

`m_As_A(\Delta T_a) = m_w s_w \Delta T_w`

`0.4(900)(115 - T) = 0.5(4186)(T - 15)`

`115 - T = 5.81(T - 15)`

`115 - T = 5.81 T - 87.2`

`T = 29.7 ^0C`

so final temperature at equilibrium will be 29.7 degree C as heat given by aluminium must be same as heat absorbed by water

Answer 2

The heat liberated by the aluminium container would be equal to heat absorbed by the water.

According to the first law of thermodynamics for a closed system the total energy of the system is always conserved. Therefore the energy liberated by the aluminium container will completely absorbed by the water untill they both come to thermal equilibrium.