Question

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy for the copper sphere. If the specific heat capacity of copper is 387 J/(kg ⋅ C°), what was the increase in temperature?

0.23 C°
0.81 C°
1.3 C°
2.1 C°

Answer 1

To calculate the increase in temperature for the copper sphere, one needs to equate its kinetic energy to the heat gained by the copper and solve for the temperature increase using the copper's specific heat capacity. Without knowing the mass of the sphere, the calculation cannot be completed.

Step-by-step explanation:

Understanding the Increase in Temperature of a Copper Sphere

When a copper sphere moving at 25 m/s hits another object and all its kinetic energy (KE) is converted into thermal energy, the increase in temperature can be calculated using the equation for kinetic energy and the formula for heat transfer involving specific heat capacity:

KE = ½ * m * v2

Where m is the mass of the sphere and v is the velocity. Once the kinetic energy is known, his is linked to the thermal energy (ΔQ) gained by the copper sphere:

ΔQ = m * c * ΔT

Here, c is the specific heat capacity of copper, and ΔT is the temperature increase we need to find. Given that the entire KE is converted to thermal energy, we set ΔQ equal to KE and solve for ΔT, which directly gives us the increase in temperature of the copper sphere.

Without the mass of the copper sphere, we cannot determine the exact temperature increase from the information provided. However, if we have the mass, we can use the above equations to find the numerical value of the increase in temperature.

Answer 2

`\Delta T = 0.81 ^oC`

Step-by-step explanation:

As we know by energy conservation

All its kinetic energy will convert into thermal energy to raise its temperature

`(1)/(2)mv^2 = ms\Delta T`

now divide both sides by mass of the object

`(1)/(2)v^2 = s\Delta T`

so change in temperature is given as

`\Delta T = (v^2)/(2s)`

`\Delta T = (25^2)/(2* 387)`

`\Delta T = 0.81^oC`