Question

lesson 31b - specific heat capacity
round to two significant figures.
you heat 47 g of gold from 195 k to 288 k

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

Explanation:

Step1: Calculate temperature change

$\Delta T = T_{final} - T_{initial} = 288\ \text{K} - 195\ \text{K} = 93\ \text{K}$

Step2: Recall specific heat of gold

Specific heat capacity of gold: $c = 0.129\ \text{J/g·K}$ (standard value)

Step3: Calculate energy in Joules

$Q = mc\Delta T = 47\ \text{g} \times 0.129\ \text{J/g·K} \times 93\ \text{K} \approx 560\ \text{J}$ (rounded to 2 sig figs)

Step4: Convert Joules to Calories

$1\ \text{Calorie} = 4184\ \text{J}$, so $Q = \frac{560\ \text{J}}{4184\ \text{J/Cal}} \approx 0.13\ \text{Cal}$ (rounded to 2 sig figs)

Answer:

1. 93 K
2. 0.129 J/g·K
3. 560 J
4. 0.13 Cal