Question

1. if an engineer’s iron ring (c = 0.449 j/g°c) has a mass of 5.5 grams and changes temperature from 25.0°c to 28.0°c, how much energy has it absorbed?

Explanation:

Step1: Calculate the temperature change

The formula for temperature change is $\Delta T=T_{final}-T_{initial}$.
Here, $T_{initial} = 25.0^{\circ}C$ and $T_{final}=28.0^{\circ}C$.
So, $\Delta T=28.0 - 25.0=3.0^{\circ}C$.

Step2: Use the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change.
Given $m = 5.5g$, $c=0.449J/g^{\circ}C$, and $\Delta T = 3.0^{\circ}C$.
Substitute the values into the formula: $Q=(5.5g)\times(0.449J/g^{\circ}C)\times(3.0^{\circ}C)$.
First, calculate $5.5\times0.449 = 2.4695$.
Then, $2.4695\times3.0=7.4085\approx7.4J$.

Answer:

$7.4J$