The energy that must be removed from a substance in order to decrease its temperature by [tex]\Delta T[/tex] is given by
[tex]Q=m C_s \Delta T[/tex]
where
m is the mass of the substance
[tex]C_s[/tex] is its specific heat capacity
[tex]\Delta T[/tex] is the variation of temperature
In our problem, m=500 g=0.5 kg, [tex]C_s = 2090 J /kg K[/tex] and the variation of temperature is
[tex]\Delta T=-25^{\circ}C-0^{\circ}C=-25^{\circ}C[/tex]
which corresponds to 25 K, since the differences in temperature are equals in Kelvin and Celsius. Therefore, the amount of energy that must be removed from the block of ice is
[tex]Q=m C_s \Delta T=(0.5 kg)(2090 J/kg K)(-25 K)=-26125 J=-2.61 \cdot 10^4 J[/tex]
where the negative sign means the energy is removed from the system.
