A 1.5 kg copper block is given an initial speed of 3.00 m/s on a rough horizontal surface. Because of friction, the block eventually comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature. (b) What happens to the remaining energy?
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(a) To calculate the increase in temperature of the copper block, we first need to determine its initial kinetic energy. The kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the block and \( v \) is its initial speed. Plugging in the given values, we get:
\[ KE = \frac{1}{2} \times 1.5 \text{ kg} \times (3.00 \text{ m/s})^2 = \frac{1}{2} \times 1.5 \times 9 = 6.75 \text{ J} \]
Now, it’s given that 85% of this energy is converted into internal energy, which is responsible for the increase in temperature. Therefore, the energy absorbed as internal energy is:
\[ \text{Internal Energy} = 0.85 \times 6.75 \text{ J} = 5.7375 \text{ J} \]
The relationship between the energy transfer as heat (\( Q \)), the mass (\( m \)), the specific heat capacity (\( c \)), and the temperature change (\( \Delta T \)) is given by:
\[ Q = mc\Delta T \]
The specific heat capacity \( c \) for copper is approximately \( 385 \text{ J/(kg·°C)} \). Solving for \( \Delta T \) and substituting the known values, we have:
\[ 5.7375 \text{ J} = 1.5 \text{ kg} \times 385 \text{ J/(kg·°C)} \times \Delta T \]
\[ \Delta T = \frac{5.7375 \text{ J}}{1.5 \text{ kg} \times 385 \text{ J/(kg·°C)}} \]
\[ \Delta T = \frac{5.7375}{577.5} \text{ °C} \]
\[ \Delta T \approx 0.00994 \text{ °C} \]
So, the increase in temperature of the copper block is approximately 0.00994 degrees Celsius.
(b) The remaining 15% of the energy is probably dissipated into the environment—mostly into the surrounding air and the surface on which the block was sliding—due to friction and as sound. These types of energy transfers typically heat the surroundings rather than the object in question.