Let’s solve each problem step by step. We’ll use the formula for heat transfer:
q = m × c × ΔT
Where:
-
q = heat energy (in Joules)
-
m = mass (in grams)
-
c = specific heat capacity (J/g°C)
-
ΔT = change in temperature = T_final - T_initial (°C)
We’ll rearrange this formula as needed to find what each question asks.
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Problem 1:
*How much heat is required to raise the temperature of 12.0 g of silver from 15.0°C to 35.0°C? Specific heat of silver = 0.24 J/g°C.*
Step 1: Find ΔT
ΔT = 35.0 – 15.0 = 20.0°C
Step 2: Plug into formula
q = m × c × ΔT
q = 12.0 g × 0.24 J/g°C × 20.0°C
q = 12.0 × 0.24 = 2.88
2.88 × 20.0 =
57.6 J
✔ Answer:
57.6 J
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Problem 2:
*How much heat, in joules, is required to raise the temperature of 1.50 kg of water by 10.0°C? Specific heat of water = 4.184 J/g°C.*
Wait — mass is given in kg, but specific heat uses grams. Convert kg to g:
1.50 kg = 1500 g
ΔT = 10.0°C (already given)
q = m × c × ΔT
q = 1500 g × 4.184 J/g°C × 10.0°C
First: 1500 × 4.184 = 6276
Then: 6276 × 10.0 =
62,760 J
✔ Answer:
62,760 J
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Problem 3:
*What is the mass of a sample of copper if it requires 298 J to raise its temperature from 25.0°C to 50.0°C? Specific heat of copper = 0.385 J/g°C.*
We need to find
m, so rearrange the formula:
q = m × c × ΔT → m = q / (c × ΔT)
ΔT = 50.0 – 25.0 = 25.0°C
m = 298 J / (0.385 J/g°C × 25.0°C)
First: 0.385 × 25.0 = 9.625
Then: 298 ÷ 9.625 ≈ ?
Let’s calculate:
9.625 × 30 = 288.75
298 – 288.75 = 9.25
So 30 + (9.25 / 9.625) ≈ 30 + 0.96 ≈
30.96 g
Rounded to 3 significant figures:
31.0 g
✔ Answer:
31.0 g
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Problem 4:
*What is the specific heat of iron if it requires 460 J to raise the temperature of 20.0 g of iron by 50.0°C?*
We need to find
c, so rearrange:
q = m × c × ΔT → c = q / (m × ΔT)
c = 460 J / (20.0 g × 50.0°C)
= 460 / 1000
=
0.460 J/g°C
✔ Answer:
0.460 J/g°C
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Problem 5:
*What is the temperature change on a 75.0 g sample of mercury if 2000 J of heat are added? Specific heat of mercury = 0.140 J/g°C.*
We need to find
ΔT, so rearrange:
q = m × c × ΔT → ΔT = q / (m × c)
ΔT = 2000 J / (75.0 g × 0.140 J/g°C)
First: 75.0 × 0.140 = 10.5
Then: 2000 ÷ 10.5 ≈ ?
10.5 × 190 = 1995
2000 – 1995 = 5 → so ≈ 190 + (5/10.5) ≈ 190.48
Rounded to 3 significant figures:
190°C
✔ Answer:
190°C
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Final Answer:
1. 57.6 J
2. 62,760 J
3. 31.0 g
4. 0.460 J/g°C
5. 190°C
Parent Tip: Review the logic above to help your child master the concept of thermochemistry worksheet.
