The Absorption of Heat by Solids and Liquids — Question 3

Question 3

A calorimeter cup must be included in an unknown-specific-heat measurement

A .200 kg hot metal at $100^\circ$C is placed in .300 kg water initially at $20.0^\circ$C. The cup also starts at $20.0^\circ$C and has heat capacity 80.0 J/K. The final temperature is $25.0^\circ$C. Use water specific heat $4200\,\mathrm{J/(kg\,K)}$; assume an insulated apparatus and no phase change.

1. Infer the metal specific heat.

2. Calculate the value obtained if cup heat capacity is ignored and quantify the percentage underestimate relative to the correct result.

Question 3 - Solution

A calorimeter cup must be included in an unknown-specific-heat measurement

(a) Method and calculation

$$m_mc_m(100-25)=(m_wc_w+C_{\mathrm{cup}})(25-20).$$$$c_m=\frac{(.300\cdot4200+80)(5)}{.200(75)}=446.667\,\mathrm{J/(kg\,K)}.$$

(b) Interpretation and further calculation

$$c_{m,\mathrm{omit}}=\frac{1260(5)}{15}=420\,\mathrm{J/(kg\,K)}.$$$$\frac{446.667-420}{446.667}\times100\%\approx5.970\%.$$The cup absorbs 400 J, which must be included in the metal's heat loss.

Physical check

The final temperature is between the initial temperatures; ignoring a heat recipient makes the inferred donor capacity too small.