Question

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?.

Answer 1

`30\; {\rm ^(\circ) C}`.

Step-by-step explanation:

Let
`T\; {\rm ^(\circ) C}` denote the equilibrium temperature of the two blocks. By the time the two blocks reached this temperature:

• Temperature of block
  `1` would have changed by
  `\Delta T_(1) = (T - 90)` (degrees kelvin.)
• Temperature of block
  `2` would have changed by
  `\Delta T_(2) = (T - 0) = T` (degrees kelvin.)

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

Let
`c_(1)` and
`c_(2)` denote the specific heat of the two blocks. Let
`m_(1)` and
`m_(2)` denote the mass of the two blocks. Energy change of the two blocks would be:

• 
  `Q_(1) = c_(1)\, m_(1)\, \Delta T_(1) = (1.0)\, (10)\, (T - 90) = 10\, T - 900` (joules) for block
  `1`.
• 
  `Q_(2) = c_(2)\, m_(2)\, \Delta T_(2) = (0.5)\, (40)\, (T) = 20\, T` (joules) for block
  `2`.

Under the assumptions, energy should be conserved. Hence, the total energy change should be
`0`. Therefore:

`Q_(1) + Q_(2) = 0`.

`(10\, T - 900) + (20\, T) = 0`.

`T = 30` (degrees celsius.)

In other words, the temperature of the two blocks would be
`30\; {\rm ^(\circ) C}` when the system reached equilibrium.