Question

a cube of iron (c = 0.450 j/g·°c) with a mass of 55.8 g is heated from 25.0°c to 49.0°c. how much heat is required for this process? round your answer to three significant figures.
use the formula q = mc, δt.
j
done

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=49.0^{\circ}C - 25.0^{\circ}C=24.0^{\circ}C$

Step2: Substitute values into the heat formula

Given $m = 55.8\ g$, $C_{p}=0.450\ J/g\cdot^{\circ}C$, and $\Delta T = 24.0^{\circ}C$.
Using the formula $q=mC_{p}\Delta T$, we have $q=(55.8\ g)\times(0.450\ J/g\cdot^{\circ}C)\times(24.0^{\circ}C)$
First, calculate $(55.8\times0.450)=25.11$. Then, $25.11\times24.0 = 602.64$

Answer:

$603\ J$