Answer:
Heat absorbed, Q = 110110 J
Explanation:
It is given that,
The specific heat of copper is, [tex]c=385\ J/kg.K[/tex]
Mass of block, m = 2.6 kg
Initial temperature, [tex]T_1=340\ K[/tex]
Final temperature, [tex]T_2=450\ K[/tex]
Thermal energy is given by :
[tex]Q=mc\Delta T[/tex]
[tex]Q=mc(T_2-T_1)[/tex]
[tex]Q=2.6\times 385\times (450-340)[/tex]
Q = 110110 J
So, the thermal heat of 110110 J is absorbed. Hence, this is the required solution.
