Specific Heat Worksheet Answers 1 - Free Printable

Let's solve each problem step by step using the formula for heat transfer:

$$
q = m \times c \times \Delta T
$$

Where:
- $ q $ = heat energy (in joules, J)
- $ m $ = mass (in grams, g)
- $ c $ = specific heat capacity (in J/g°C)
- $ \Delta T $ = change in temperature = $ T_{\text{final}} - T_{\text{initial}} $ (in °C)

---

Example:

50 g of gold with a specific heat of 0.129 J/g°C is heated to 115°C; it cools until the final temperature is 29.3°C. Calculate the heat of the metal.

We are calculating the heat lost as it cools.

$$
\Delta T = T_{\text{initial}} - T_{\text{final}} = 115^\circ C - 29.3^\circ C = 85.7^\circ C
$$

Now use:
$$
q = m \times c \times \Delta T = 50 \, \text{g} \times 0.129 \, \text{J/g°C} \times 85.7^\circ C
$$

$$
q = 50 \times 0.129 \times 85.7 = 561.4 \, \text{J}
$$

So, the heat released (since it's cooling) is:
$$
q_{\text{metal}} = \boxed{561.4} \, \text{J}
$$

> Note: The sign would be negative if we were being precise about direction (exothermic), but since the question asks for "heat of the metal", and doesn't specify direction, we report magnitude.

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## Problem 1:
Ethanol has a specific heat of 2.44 J/g°C. The temperature of 34.4 g of ethanol increases from 25°C to 78.8°C. How much heat was absorbed?

Given:
- $ m = 34.4 \, \text{g} $
- $ c = 2.44 \, \text{J/g°C} $
- $ \Delta T = 78.8^\circ C - 25^\circ C = 53.8^\circ C $

$$
q = m \times c \times \Delta T = 34.4 \times 2.44 \times 53.8
$$

First calculate:
$$
34.4 \times 2.44 = 83.936
$$
Then:
$$
83.936 \times 53.8 \approx 4513.5 \, \text{J}
$$

So,
$$
q = \boxed{4514} \, \text{J} \quad \text{(rounded to nearest whole number)}
$$

✔ Answer: 4514 J of heat was absorbed.

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## Problem 2:
3 grams of Aluminum was heated from 20°C to 662°C while absorbing 1728 J of heat. What is the specific heat of Aluminum?

Use:
$$
q = m \times c \times \Delta T \Rightarrow c = \frac{q}{m \times \Delta T}
$$

Given:
- $ q = 1728 \, \text{J} $
- $ m = 3 \, \text{g} $
- $ \Delta T = 662^\circ C - 20^\circ C = 642^\circ C $

$$
c = \frac{1728}{3 \times 642} = \frac{1728}{1926} \approx 0.897 \, \text{J/g°C}
$$

Compare to table value: Aluminum is listed as 0.902 J/g°C, so this is very close.

✔ Answer: $ c = \boxed{0.897} \, \text{J/g°C} $

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## Problem 3:
A metal mass of 5 grams absorbs 240 J of heat. Temperature increases by 200°C. What is the specific heat?

Given:
- $ q = 240 \, \text{J} $
- $ m = 5 \, \text{g} $
- $ \Delta T = 200^\circ C $

$$
c = \frac{q}{m \times \Delta T} = \frac{240}{5 \times 200} = \frac{240}{1000} = 0.24 \, \text{J/g°C}
$$

✔ Answer: $ c = \boxed{0.24} \, \text{J/g°C} $

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✔ Final Answers:

1. Heat absorbed by ethanol: 4514 J
2. Specific heat of aluminum: 0.897 J/g°C
3. Specific heat of unknown metal: 0.24 J/g°C

Let me know if you'd like these written out neatly on the worksheet format!