Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
Q = mcΔT
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
The thermal energy loose by the aluminum block will be Q = 22500 J
It is given that Given data,
The mass of the aluminum block, m = 1.0 kg
The initial temperature of the block, T = 50° C
Since the temperature of the block is halved then,
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C = 25° C
The specific heat capacity of aluminum,
c = 900 J/kg°C
The formula to find out the thermal energy will be
[tex]Q=m\times c\times\Delta T[/tex]
[tex]Q=1.0\times 900\times 25[/tex]
[tex]Q= 22500J[/tex]
Thus the thermal energy loose by the aluminum block will be Q = 22500 J
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