Question

lesson 31b - specific heat capacity
round to two significant figures.
you heat 670 g of iron from 201 k to 265 k.

1. what is the change in temperature?
2. what is the specific heat capacity of iron?
3. how much energy will be transferred in joules?
4. how much will be transferred in calories?

Explanation:

Step1: Calculate temperature change

$\Delta T = 265\ \text{K} - 201\ \text{K} = 64\ \text{K}$

Step2: Recall specific heat of iron

$c_{\text{iron}} = 0.45\ \text{J/g·K}$ (standard value)

Step3: Calculate energy in Joules

$Q = mc\Delta T = 670\ \text{g} \times 0.45\ \text{J/g·K} \times 64\ \text{K} = 19584\ \text{J} \approx 2.0 \times 10^4\ \text{J}$

Step4: Convert Joules to Calories

$1\ \text{Calorie} = 4184\ \text{J}$, so $Q = \frac{19584\ \text{J}}{4184\ \text{J/Calorie}} \approx 4.7\ \text{Calories}$

Answer:

1. 64 K
2. 0.45 J/g·K
3. $2.0 \times 10^4$ J
4. 4.7 Calories