Question

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in this process, in joules?

Answer 1

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$

Initial temperature, $T_i=88^{\circ}C$

Final temperature, $T_f=45^{\circ}C$

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

$Q=mc\Delta T$

Let the mass of the object is 10 g or 0.01 kg

So,

$Q=0.01\times 345\times (88-45)$

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

Answer 2

Answer:

14835 J

Explanation:

Recall that heat capacity, C, is the ratio of heat to the change in temperature.

                     C= q/ΔT

Rewrite the equation to solve for q.

                       q=CΔT

In order to solve for q, you must first calculate ΔT.

                       ΔT=Tfinal−Tinitial

                       ΔT=45.0∘C–88.0∘C=−43.0∘C

Next, Substitute the known values into the equation above and solve for q.

                      q=(345J∘C)(−43.0∘C)= −14,835 J

Round your answer to the nearest hundred to use correct significant figures.

                    q= −14,800 J

The negative sign means that heat is released, or the process is exothermic. Therefore, the amount of heat released is about 14,800 J.

Hope this helped!