The Absorption of Heat by Solids and Liquids — Question 1

Section 18-7 · Physics pages 2932–2933

Question 1 A composite sample has additive heat capacity A bonded assembly contains .400 kg of material A with cA = 900 J/(kg K) and .600 kg of B with cB = 400 J/(kg K). The parts remain at one temperature. Neglect bonding-layer capacity, losses and work; both specific heats are constant. A: .400 kg B: .600 kg common temperature Schematic; labels, not drawing lengths, specify the data. (a) Find the assembly heat capacity and heat required for a 25.0 K rise. (b) Find its mass-averaged specific heat. How much material B must instead be used with the same .400 kg of A to make the assembly heat capacity 1000 J/K?
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Question 1 - Solution A composite sample has additive heat capacity A: .400 kg B: .600 kg common temperature (a) Method and calculation C = mAcA + mBcB = .400(900) + .600(400) = 600 J/K. Q = C∆T = 600(25) = 15000 J. (b) Interpretation and further calculation ceff= C mA + mB = 600 J/(kg K). 360 + 400mB = 1000 ⇒mB = 1.60 kg. An unweighted average of the two specific heats would give 650 and is inappropriate for unequal masses. Physical check Heat capacity adds directly; specific heat must be weighted by mass.

Question 1

A composite sample has additive heat capacity

A bonded assembly contains .400 kg of material A with cA=900J/(kgK)c_A=900\,\mathrm{J/(kg\,K)} and .600 kg of B with cB=400J/(kgK)c_B=400\,\mathrm{J/(kg\,K)}. The parts remain at one temperature. Neglect bonding-layer capacity, losses and work; both specific heats are constant.

Diagram shown in the original SVG worksheet.

  1. Find the assembly heat capacity and heat required for a 25.0 K rise.

  2. Find its mass-averaged specific heat. How much material B must instead be used with the same .400 kg of A to make the assembly heat capacity 1000 J/K?

Question 1 - Solution

A composite sample has additive heat capacity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

C=mAcA+mBcB=.400(900)+.600(400)=600J/K.C=m_Ac_A+m_Bc_B=.400(900)+.600(400)=600\,\mathrm{J/K}.Q=CΔT=600(25)=15000J.Q=C\Delta T=600(25)=15000\,\mathrm J.

(b) Interpretation and further calculation

ceff=CmA+mB=600J/(kgK).c_{\mathrm{eff}}=\frac{C}{m_A+m_B}=600\,\mathrm{J/(kg\,K)}.360+400mB=1000⇒mB=1.60kg.360+400m_B=1000\ \Rightarrow\ m_B=1.60\,\mathrm{kg}.An unweighted average of the two specific heats would give 650 and is inappropriate for unequal masses.

Physical check

Heat capacity adds directly; specific heat must be weighted by mass.