# How to find the temperature variation after a collision between two blocks?

#### Chemist116

The problem is as follows:

Two blocks are shown in the figure from below. Both have the same mass and are at $28^{\circ}C$. After the collision both are stick together and only absorb 30% of the dissipated energy which acts to increase their temperature. Using this information find the change in their temperature (in $10^{-3}\,^{\circ}C$) of the blocks. You may consider that the $ce_{block}=0.2\frac{cal}{g\cdot ^{\circ}C}$ and $1J=0.24\,\textrm{cal}$

The alternatives are as follows:

$\begin{array}{ll} 1.&80\\ 2.&40\\ 3.&0.68\\ 4.&0.18\\ \end{array}$

What I've attempted here was to use the conservation of momentum as follows:

$p_i=p_f$

Then

$5m+3m=2mu$

$u=4$

Therefore the final kinetic energy is:

$E_k=\frac{1}{2}(2m)(4^2)=16m\,J$

The initial kinetic energy would be the combined kinetic energies of both blocks:

$E_{k1}+E_{k2}=\frac{1}{2}m(5^2)+\frac{1}{2}m(3^2)=17m\,J$

Then the absorbed energy must be difference:

$17m-16m=1m$

From this energy only $30\%$ is used to increase the temperature of the blocks and considering the units from Joules to calories, hence:

$(0.3)(1m)(0.24) = mc_{e}(\Delta T)$

Here the mass is expressed in Kilograms so to be consistent the specific heat is expressed in terms of cal per kilograms celcius.

$(0.3)(1m)(0.24) = mc_{e}\times(10^3)(\Delta T)$

$(0.3)(1)(0.24)=0.2\times 10^3 \Delta T$

$\Delta T = 0.36 \times 10^{-3}$

But supposedly the answer is the fifth alternative according to my book. What could it be what's not right here?. Any help please?.