Question

how much j of heat is required to increase the temperature of 3.51 kg of iron from 25.00 °c to 147 °c? the specific heat capacity of iron is 0.451 j/g·°c. solve to 3 significant figures.

Explanation:

Step1: Recall the heat formula

The formula for heat \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Convert mass to grams

Given \( m = 3.51\space kg = 3.51\times1000 = 3510\space g \).

Step3: Calculate temperature change

\( \Delta T = 147^{\circ}C - 25.00^{\circ}C = 122^{\circ}C \).

Step4: Substitute values into the formula

\( q = 3510\space g\times0.451\space J/g^{\circ}C\times122^{\circ}C \).
First, calculate \( 3510\times0.451 = 1583.01 \).
Then, \( 1583.01\times122 = 193127.22\space J \).

Step5: Round to 3 significant figures

\( 193127.22\space J \approx 1.93\times10^{5}\space J \) (or \( 193000\space J \) when rounded to 3 significant figures).

Answer:

\( 1.93\times10^{5}\space J \) (or \( 193000\space J \))