Calculating Specific Heat Worksheet Q = mc∆T, where Q = heat energy, m = mass, and ∆T = change in temperature.

Calculating Specific Heat Worksheet Q = mc∆T, where Q = heat energy, m = mass, and ∆T = change in temperature. Remember, ∆T = (Tfinal – Tinitial). Show all work and proper units. 1) A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron. 2) How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C? 3) Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 67,500 joules of heat, and its temperature changes from 32°C to 57°C. 4) 100.0 g of 4.0°C water is heated until its temperature is 37°C. If the specific heat of water is 4.18 J/g°C, calculate the amount of heat energy needed to cause this rise in temperature. 5) 25.0 g of mercury is heated from 25°C to 155°C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury. 6) What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3 J of heat and the temperature rises 15.0°C? 7) What mass of water will change its temperature by 30°C when 525 J of heat is added to it? The specific heat of water is 4.18 J/g°C.

The Correct Answer and Explanation is:

Formula:Q=mcΔTQ = mc\Delta TQ=mcΔT

Where:

• QQQ = heat energy (joules),
• mmm = mass (grams),
• ccc = specific heat capacity (J/g°C),
• ΔT\Delta TΔT = change in temperature (°C), where ΔT=Tfinal−Tinitial\Delta T = T_{final} – T_{initial}ΔT=Tfinal​−Tinitial​.

1) Specific Heat Capacity of Iron

Given:

• Mass (mmm) = 15.75 g,
• Heat energy (QQQ) = 1086.75 J,
• Initial temperature (TinitialT_{initial}Tinitial​) = 25°C,
• Final temperature (TfinalT_{final}Tfinal​) = 175°C.

First, find ΔT\Delta TΔT:ΔT=Tfinal−Tinitial=175°C−25°C=150°C\Delta T = T_{final} – T_{initial} = 175°C – 25°C = 150°CΔT=Tfinal​−Tinitial​=175°C−25°C=150°C

Now, solve for ccc (specific heat):1086.75 J=15.75 g×c×150°C1086.75 \, J = 15.75 \, g \times c \times 150°C1086.75J=15.75g×c×150°Cc=1086.7515.75×150=0.459 J/g°Cc = \frac{1086.75}{15.75 \times 150} = 0.459 \, J/g°Cc=15.75×1501086.75​=0.459J/g°C

2) Heat Energy for Aluminum

Given:

• Mass (mmm) = 10.0 g,
• Specific heat of aluminum (ccc) = 0.90 J/g°C,
• Initial temperature (TinitialT_{initial}Tinitial​) = 22°C,
• Final temperature (TfinalT_{final}Tfinal​) = 55°C.

Find ΔT\Delta TΔT:ΔT=55°C−22°C=33°C\Delta T = 55°C – 22°C = 33°CΔT=55°C−22°C=33°C

Now, solve for QQQ (heat energy):Q=m×c×ΔT=10.0 g×0.90 J/g°C×33°CQ = m \times c \times \Delta T = 10.0 \, g \times 0.90 \, J/g°C \times 33°CQ=m×c×ΔT=10.0g×0.90J/g°C×33°CQ=297 JQ = 297 \, JQ=297J

3) Specific Heat Capacity of Wood

Given:

• Mass (mmm) = 1500.0 g,
• Heat energy (QQQ) = 67,500 J,
• Initial temperature (TinitialT_{initial}Tinitial​) = 32°C,
• Final temperature (TfinalT_{final}Tfinal​) = 57°C.

Find ΔT\Delta TΔT:ΔT=57°C−32°C=25°C\Delta T = 57°C – 32°C = 25°CΔT=57°C−32°C=25°C

Now, solve for ccc (specific heat):67,500 J=1500.0 g×c×25°C67,500 \, J = 1500.0 \, g \times c \times 25°C67,500J=1500.0g×c×25°Cc=67,5001500.0×25=1.80 J/g°Cc = \frac{67,500}{1500.0 \times 25} = 1.80 \, J/g°Cc=1500.0×2567,500​=1.80J/g°C

4) Heat Energy for Water

Given:

• Mass (mmm) = 100.0 g,
• Specific heat of water (ccc) = 4.18 J/g°C,
• Initial temperature (TinitialT_{initial}Tinitial​) = 4.0°C,
• Final temperature (TfinalT_{final}Tfinal​) = 37°C.

Find ΔT\Delta TΔT:ΔT=37°C−4.0°C=33°C\Delta T = 37°C – 4.0°C = 33°CΔT=37°C−4.0°C=33°C

Now, solve for QQQ (heat energy):Q=m×c×ΔT=100.0 g×4.18 J/g°C×33°CQ = m \times c \times \Delta T = 100.0 \, g \times 4.18 \, J/g°C \times 33°CQ=m×c×ΔT=100.0g×4.18J/g°C×33°CQ=13,794 JQ = 13,794 \, JQ=13,794J

5) Specific Heat Capacity of Mercury

Given:

• Mass (mmm) = 25.0 g,
• Heat energy (QQQ) = 455 J,
• Initial temperature (TinitialT_{initial}Tinitial​) = 25°C,
• Final temperature (TfinalT_{final}Tfinal​) = 155°C.

Find ΔT\Delta TΔT:ΔT=155°C−25°C=130°C\Delta T = 155°C – 25°C = 130°CΔT=155°C−25°C=130°C

Now, solve for ccc (specific heat):455 J=25.0 g×c×130°C455 \, J = 25.0 \, g \times c \times 130°C455J=25.0g×c×130°Cc=45525.0×130=0.141 J/g°Cc = \frac{455}{25.0 \times 130} = 0.141 \, J/g°Cc=25.0×130455​=0.141J/g°C

6) Specific Heat Capacity of Silver

Given:

• Mass (mmm) = 55.00 g,
• Heat energy (QQQ) = 47.3 J,
• Temperature rise (ΔT\Delta TΔT) = 15.0°C.

Now, solve for ccc (specific heat):47.3 J=55.00 g×c×15.0°C47.3 \, J = 55.00 \, g \times c \times 15.0°C47.3J=55.00g×c×15.0°Cc=47.355.00×15.0=0.0574 J/g°Cc = \frac{47.3}{55.00 \times 15.0} = 0.0574 \, J/g°Cc=55.00×15.047.3​=0.0574J/g°C

7) Mass of Water with Heat Energy

Given:

• Heat energy (QQQ) = 525 J,
• Specific heat of water (ccc) = 4.18 J/g°C,
• Temperature change (ΔT\Delta TΔT) = 30°C.

Now, solve for mass (mmm):Q=m×c×ΔTQ = m \times c \times \Delta TQ=m×c×ΔT525 J=m×4.18 J/g°C×30°C525 \, J = m \times 4.18 \, J/g°C \times 30°C525J=m×4.18J/g°C×30°Cm=5254.18×30=4.19 gm = \frac{525}{4.18 \times 30} = 4.19 \, gm=4.18×30525​=4.19g

Summary of Results:

1. Specific heat of iron = 0.459 J/g°C
2. Heat needed for aluminum = 297 J
3. Specific heat of wood = 1.80 J/g°C
4. Heat energy for water = 13,794 J
5. Specific heat of mercury = 0.141 J/g°C
6. Specific heat of silver = 0.0574 J/g°C
7. Mass of water = 4.19 g