Two blocks are enclosed in a perfectly insulated box. Block 1 at 90 c has mass of 10 kg, specific heat of 1. 0 j/k/m. Block 2 at 0 c has a mass of 40 kg, specific heat of 0. 5 j/k/m. What is the final temperature of the two blocks in equilibrium?.

Temperature of block [tex]1[/tex] would have changed by [tex]\Delta T_{1} = (T - 90)[/tex] (degrees kelvin.)
Temperature of block [tex]2[/tex] would have changed by [tex]\Delta T_{2} = (T - 0) = T[/tex] (degrees kelvin.)

(Note that the temperature change of one degree celsius is equivalent to a temperature change of one degree kelvin.)

Let [tex]c_{1}[/tex] and [tex]c_{2}[/tex] denote the specific heat of the two blocks. Let [tex]m_{1}[/tex] and [tex]m_{2}[/tex] denote the mass of the two blocks. Energy change of the two blocks would be:

[tex]Q_{1} = c_{1}\, m_{1}\, \Delta T_{1} = (1.0)\, (10)\, (T - 90) = 10\, T - 900[/tex] (joules) for block [tex]1[/tex].
[tex]Q_{2} = c_{2}\, m_{2}\, \Delta T_{2} = (0.5)\, (40)\, (T) = 20\, T[/tex] (joules) for block [tex]2[/tex].

Under the assumptions, energy should be conserved. Hence, the total energy change should be [tex]0[/tex]. Therefore:

[tex]Q_{1} + Q_{2} = 0[/tex].

[tex](10\, T - 900) + (20\, T) = 0[/tex].

[tex]T = 30[/tex] (degrees celsius.)

In other words, the temperature of the two blocks would be [tex]30\; {\rm ^{\circ} C}[/tex] when the system reached equilibrium.