QuestionAugust 22, 2025

4. A 50 gram piece of iron warms from 0^circ C to 40^circ C while absorbing 900 joules.What is the specific heat of iron?

4. A 50 gram piece of iron warms from 0^circ C to 40^circ C while absorbing 900 joules.What is the specific heat of iron?
4. A 50 gram piece of iron warms from 0^circ C to 40^circ C while absorbing 900 joules.What is
the specific heat of iron?

Solution
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Answer

450 \text{ J/kg}^{\circ}\text{C} Explanation 1. Identify the formula for specific heat Use the formula **q = mc\Delta T**, where q is heat absorbed, m is mass, c is specific heat, and \Delta T is the change in temperature. 2. Rearrange the formula to solve for specific heat Rearrange to **c = \frac{q}{m\Delta T}**. 3. Substitute the given values q = 900 \text{ J}, m = 50 \text{ g} = 0.05 \text{ kg}, \Delta T = 40^{\circ}C - 0^{\circ}C = 40^{\circ}C. 4. Calculate the specific heat c = \frac{900}{0.05 \times 40}. 5. Perform the calculation c = \frac{900}{2} = 450 \text{ J/kg}^{\circ}\text{C}.

Explanation

1. Identify the formula for specific heat<br /> Use the formula **$q = mc\Delta T$**, where $q$ is heat absorbed, $m$ is mass, $c$ is specific heat, and $\Delta T$ is the change in temperature.<br /><br />2. Rearrange the formula to solve for specific heat<br /> Rearrange to **$c = \frac{q}{m\Delta T}$**.<br /><br />3. Substitute the given values<br /> $q = 900 \text{ J}$, $m = 50 \text{ g} = 0.05 \text{ kg}$, $\Delta T = 40^{\circ}C - 0^{\circ}C = 40^{\circ}C$.<br /><br />4. Calculate the specific heat<br /> $c = \frac{900}{0.05 \times 40}$.<br /><br />5. Perform the calculation<br /> $c = \frac{900}{2} = 450 \text{ J/kg}^{\circ}\text{C}$.
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