Question

if you pour equal amounts of scalding hot water into different metallic cups of equal temperature, which cup will heat up the most? rank the cups from hottest to coldest.
rank from warmest to coolest final temperature.
view available hint(s)
500 - g silver cup 750 - g aluminum cup 500 - g aluminum cup
warmest coolest

Explanation:

Step1: Recall the formula for heat transfer

The heat transfer formula is \(Q = mc\Delta T\), where \(Q\) is the heat transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. For the water - cup system (assuming no heat loss to the environment), \(Q_{water}=-Q_{cup}\). Let the initial temperature of water be \(T_{w,i}\), initial temperature of cups be \(T_{c,i}\), and final temperature be \(T_f\). Then \(m_{w}c_{w}(T_f - T_{w,i})=-m_{c}c_{c}(T_f - T_{c,i})\). Since \(T_{w,i}\) (temperature of hot water) and \(m_{w}\) (mass of water) are the same for all cases, and \(T_{c,i}\) (initial temperature of cups) is the same. We can rewrite the formula as \(T_f=\frac{m_{w}c_{w}T_{w,i}+m_{c}c_{c}T_{c,i}}{m_{w}c_{w}+m_{c}c_{c}}\). The specific heat capacity of silver \(c_{Ag}=0.24\space J/(g\cdot^{\circ}C)\) and of aluminum \(c_{Al}=0.90\space J/(g\cdot^{\circ}C)\)

Step2: Analyze the effect of mass and specific heat capacity on final temperature

• For the \(500 - g\) silver cup: \(m = 500g\), \(c = 0.24\space J/(g\cdot^{\circ}C)\). The product \(mc=500\times0.24 = 120\space J/^{\circ}C\)
• For the \(750 - g\) aluminum cup: \(m = 750g\), \(c = 0.90\space J/(g\cdot^{\circ}C)\). The product \(mc=750\times0.90=675\space J/^{\circ}C\)
• For the \(500 - g\) aluminum cup: \(m = 500g\), \(c = 0.90\space J/(g\cdot^{\circ}C)\). The product \(mc = 500\times0.90=450\space J/^{\circ}C\)

Since \(T_f=\frac{m_{w}c_{w}T_{w,i}+m_{c}c_{c}T_{c,i}}{m_{w}c_{w}+m_{c}c_{c}}\), and \(m_{w}c_{w}T_{w,i}\) and \(T_{c,i}\) are constant. A larger \(m_{c}c_{c}\) will make \(T_f\) closer to \(T_{w,i}\) (because the denominator \(m_{w}c_{w}+m_{c}c_{c}\) and numerator \(m_{w}c_{w}T_{w,i}+m_{c}c_{c}T_{c,i}\) change. If we assume \(T_{c,i}\ll T_{w,i}\), \(T_f\approx\frac{m_{w}c_{w}T_{w,i}+m_{c}c_{c}T_{c,i}}{m_{w}c_{w}+m_{c}c_{c}}\approx\frac{m_{w}c_{w}T_{w,i}}{m_{w}c_{w}+m_{c}c_{c}}+\frac{m_{c}c_{c}T_{c,i}}{m_{w}c_{w}+m_{c}c_{c}}\). The larger \(m_{c}c_{c}\), the larger \(T_f\)

Answer:

Warmest: \(750 - g\) aluminum cup, Middle: \(500 - g\) aluminum cup, Coolest: \(500 - g\) silver cup