Question

lesson 31b - specific heat capacity
round to two significant figures.
you heat 25 g of water from 185 k to 227 k.

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

Explanation:

Step1: Calculate temperature change

$\Delta T = 227\ \text{K} - 185\ \text{K} = 42\ \text{K}$

Step2: Recall specific heat of water

$c = 4.184\ \text{J/g·K}$ (standard value)

Step3: Calculate energy in Joules

$Q = mc\Delta T = 25\ \text{g} \times 4.184\ \text{J/g·K} \times 42\ \text{K} = 4393.2\ \text{J} \approx 4400\ \text{J}$ (rounded to 2 sig figs)

Step4: Convert Joules to Calories

$1\ \text{Calorie} = 4184\ \text{J}$, so $Q = \frac{4393.2\ \text{J}}{4184\ \text{J/Cal}} \approx 1.05\ \text{Cal} \approx 1.1\ \text{Cal}$ (rounded to 2 sig figs)

Answer:

1. 42 K
2. 4.184 J/g·K
3. 4400 J
4. 1.1 Calories