Specific Heat and q = mcΔT比热容与 q = mcΔT

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 6~35 min

Learning goals

  • Explain what specific heat means, and compare how much different substances warm up when they absorb the same heat
  • Use q = mcΔT to find heat, temperature change, specific heat, or mass, with correct signs
  • Calculate the heat absorbed or released during a change of state

Before you start

  • Changes of state (1.2.2)
  • Energy, heat, and temperature (2.10.1)

Why is the beach cooler than inland in summer?

At noon in summer the sand is too hot to walk on barefoot, yet the ocean is still cool. At night the sand cools quickly while the water stays mild. Sand and water sit under the same sun all day, but different substances need different amounts of heat to warm up by the same number of degrees.

Specific heat

When a substance absorbs heat, its temperature rises. The heat needed to raise the temperature of 1 g of a substance by 1 °C is its specific heat capacity比热容, or just specific heat, c, usually in J/(g·°C). Water’s specific heat is 4.18 J/(g·°C): 1 g of water absorbs 4.18 J to warm by 1 °C.

So the heat absorbed or released by a substance is

q=mcΔTq = mc\Delta T

where m is the mass and ΔT = Tfinal − Tinitial. When the temperature rises, ΔT > 0 and q > 0: the substance absorbs heat. When it falls, ΔT < 0 and q < 0: the substance releases heat. A change of 1 °C equals a change of 1 K, so ΔT is the same in either unit.

The heat needed to warm a whole object by 1 °C is its heat capacity热容, mc, which depends on how big the object is. Specific heat depends only on the substance and its state.

Specific heats of common substances

Values at 25 °C:

Substancec (J/(g·°C))
water (liquid)4.18
aluminum0.897
iron0.449
zinc0.388
copper0.385
silver0.235
gold, lead0.129

Water’s specific heat is far larger than any metal’s because of hydrogen bonding (2.4.8). Much of the energy water absorbs goes into stretching and breaking hydrogen bonds rather than speeding up the molecules. So water warms and cools slowly, which is why engines are cooled with water and coastal temperatures swing less than inland ones.

Heat in changes of state

On the heating curve in 1.2.2, the temperature stays constant while ice melts or water boils: the heat goes into overcoming the attractions between particles, not into raising the temperature. So q = mcΔT doesn’t apply. The heat for a change of state is calculated per mole:

q=n ΔHphase changeq = n\,\Delta H_{\text{phase change}}

Here n is the amount in moles, and ΔHphase change is the heat absorbed or released when 1 mol of the substance changes state, in kJ/mol (see lesson 2.10.2 for what kJ/mol means).

For water, the enthalpy of fusion熔化焓 ΔHfus is 6.01 kJ/mol at 0 °C and the enthalpy of vaporization汽化焓 ΔHvap is 40.7 kJ/mol at 100 °C. Vaporization takes far more energy than melting: it pulls the molecules completely apart, while melting only lets them slide past each other.

First compare: given the same heat, which warms most, 100 g of water, aluminum, or copper? Then switch to “Solve” and find any one quantity from the other four:

Try it: specific heat and q = mcΔT
Substances (up to three, 100 g each)

How Chinese textbooks differ

  • Chinese physics textbooks give specific heat in J/(kg·°C), with water at 4.2 × 10³ J/(kg·°C). Chinese chemistry uses J/(g·°C) with 4.18, as here. The two differ only by the factor between g and kg.
  • Chinese physics calls the heats of melting and boiling 熔化热 and 汽化热 and often gives them in J/kg. US chemistry uses ΔHfus and ΔHvap in kJ/mol.

Example 1: heating a kettle of water

How much heat does it take to warm 250.0 g of water from 20.0 °C to 100.0 °C?

Step-by-step solution

  1. The formula. q = mcΔT. For water, c = 4.18 J/(g·°C); the g in the unit means the mass goes in grams.

  2. Temperature change. Final minus initial:

    ΔT=100.0−20.0=80.0 °C\Delta T = 100.0 - 20.0 = 80.0\ \text{°C}

  3. Substitute.

    q=mcΔT=250.0×4.18×80.0=8.36×104 J\begin{aligned} q &= mc\Delta T \\ &= 250.0 \times 4.18 \times 80.0 \\ &= 8.36 \times 10^{4}\ \text{J} \end{aligned}

  4. In kJ. q = 83.6 kJ, positive because the water absorbs heat.

  5. Compare with boiling. Boiling all of that water away would take another

    n=250.0/18.02=13.87 molq=13.87×40.7≈565 kJ\begin{aligned} n &= 250.0 / 18.02 \\ &= 13.87\ \text{mol} \\ q &= 13.87 \times 40.7 \approx 565\ \text{kJ} \end{aligned}

    much more than warming it.

  6. A common mistake. Using the final temperature, 100.0 °C, as ΔT, which gives 104.5 kJ. ΔT is the change in temperature.

Example 2: identifying a metal

A 45.0 g piece of metal absorbs 1.00 × 10³ J of heat, and its temperature rises from 25.0 °C to 74.5 °C. What metal could it be?

Step-by-step solution

  1. The plan. Specific heat is a property of the substance, so find c and look it up. Rearranging q = mcΔT gives c = q ÷ (mΔT).

  2. Temperature change.

    ΔT=74.5−25.0=49.5 °C\Delta T = 74.5 - 25.0 = 49.5\ \text{°C}

  3. Specific heat.

    c=qmΔT=1.00×10345.0×49.5=0.449 J/(g⋅°C)\begin{aligned} c &= \frac{q}{m\Delta T} \\ &= \frac{1.00 \times 10^{3}}{45.0 \times 49.5} \\ &= \boxed{0.449\ \text{J/(g·°C)}} \end{aligned}

    Units: J ÷ (g × °C) = J/(g·°C).

  4. Compare with the table. That matches iron, 0.449 J/(g·°C).

  5. Think about it. Metals have specific heats about 1/9 of water’s. The same 1000 J would warm 45 g of water by only about 5 °C, but it warmed this iron by almost 50 °C.

Practice: how much heat, or how big a temperature change? Use c = 4.18 J/(g·°C) for water.
  1. Warming 10 g of water by 10 °C
  2. Warming 20 g of water by 10 °C
  3. Warming 100 g of water by 10 °C
  4. Warming 1 kg of water by 1 °C
  5. Temperature rise when 100 g of water absorbs 418 J
  6. Temperature rise when 50 g of water absorbs 418 J
  7. Temperature rise when 10 g of water absorbs 418 J
  8. Temperature rise when 200 g of water absorbs 836 J

Common mistakes

Mistake 1: “A substance with a high specific heat warms up quickly.” It’s the opposite. A high specific heat means a smaller temperature rise for the same heat. Water has a high specific heat, so it warms and cools slowly.

Mistake 2: “ΔT in °C and ΔT in K give different answers.” A change of 1 °C is a change of 1 K, so ΔT is the same either way. Temperatures themselves are different: 20 °C is not 20 K.

Mistake 3: “Heat is always positive, so ΔT is always positive.” By convention ΔT = Tfinal − Tinitial. When something releases heat its temperature drops, so ΔT < 0 and q < 0.

Quick check

Equal masses of copper and aluminum absorb the same amount of heat. Which warms up more?

50.0 g of water cools from 25.0 °C to 15.0 °C. What is q for the water?

Summary

  • Specific heat c is the heat needed to warm 1 g of a substance by 1 °C. Water’s is 4.18 J/(g·°C), much larger than any metal’s.
  • q = mcΔT with ΔT = Tfinal − Tinitial: q > 0 when heat is absorbed and q < 0 when it is released. The higher the specific heat, the smaller the temperature change for the same heat.
  • During a change of state the temperature is constant and q = nΔH. For water, ΔHfus = 6.01 kJ/mol and ΔHvap = 40.7 kJ/mol.
On to AP (for later)

The AP equation sheet gives q = mcΔT (topic 6.4). The specific heat of water isn’t on the sheet, so problems give it, usually as 4.18 J/(g·°C). Topic 6.5 asks you to calculate the heat for a phase change (q = nΔH) and to explain with intermolecular forces why ΔHvap is larger than ΔHfus. A common question asks for the heat along each segment of a heating curve and then the total.

Data sources: specific heats and enthalpies of fusion and vaporization from the CRC Handbook.

Practice this chapter: 2.10 · Thermochemistry · 10 questions

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