Question

what is the change in heat of a 1.3 - gram block of sodium when its temperature decreases from - 147°c to - 255°c?
the specific heat of sodium is 1.23 j/g°c.
round your answer to the correct number of significant figures.

joules
complete the sentence.
this process is because heat is by the sodium as it cools.

Explanation:

Step1: Calculate the temperature change

The formula for temperature change $\Delta T=T_{final}-T_{initial}$. Here, $T_{initial}=- 147^{\circ}C$ and $T_{final}=-255^{\circ}C$. So, $\Delta T=-255 - (-147)=-255 + 147=-108^{\circ}C$.

Step2: Use the heat - change formula

The formula for heat change is $Q = mc\Delta T$, where $m = 1.3g$, $c = 1.23J/g^{\circ}C$, and $\Delta T=-108^{\circ}C$.
Substitute the values: $Q=(1.3)(1.23)(-108)$.
First, calculate $(1.3)(1.23)=1.599$.
Then, $Q = 1.599\times(-108)=-172.692$.

Answer:

• 170 joules (rounded to two significant figures).

This process is exothermic because heat is released by the sodium as it cools.