1. The problem statement, all variables and given/known data An ordinary glass is filled to the brim with water at 100oc. How much water could be added to the glass if the temperature is lowered to 20oC? Assume that the coefficient of volume expansion for glass is 2.7 x 10-5 K-1 and for water it is 2.1 x 10-4 K-1. 2. Relevant equations ΔV = γ⋅VΔθ 3. The attempt at a solution Δθ = 80 K I will assume 1m3 for volume to make the calculations simple for both. ΔVglass = 2.7 x 10-5 K-1⋅1m3⋅80k = 0.00216m3 ΔVwater = 2.1 x 10-4 K-1 ⋅ 1m3⋅80k = 0.0168m3 I reasoned that since there is differential shrinkage, the glass shrinks 0.00216m3, thus there is less available room for storing water, but the water shrank also, if we subtracted the results from each other, we would have the net change. So we could add 0.01464m3 more water. I wanted to check my supposition that I could use 1m3 for both the glass and the water. The glass does not occupy a full cubic meter, since its mostly walls with a void in side. Is that right? Or does that not get addressed in intro physics? Have I worked out the problem correctly or ? I probably should report this in percentage as 1.46% change, no? Thanks!