Answer :
Answer:
[tex]30\; {\rm ^{\circ} C}[/tex].
Explanation:
Let [tex]T\; {\rm ^{\circ} C}[/tex] denote the equilibrium temperature of the two blocks. By the time the two blocks reached this temperature:
- 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.