Testing the mathjax engine - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: Not HP Calculators (/forum-7.html) +--- Forum: Test forum (/forum-18.html) +--- Thread: Testing the mathjax engine (/thread-6224.html) Testing the mathjax engine - emece67 - 05-08-2016 08:56 PM From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$ Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units. RE: Testing the mathjax engine - Csaba Tizedes - 02-19-2017 03:33 PM $\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i\cdotN_{i-1}+D_{i-1}}{D_{i-1}\cdotF_i}$ RE: Testing the mathjax engine - Csaba Tizedes - 02-19-2017 03:35 PM (05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$ Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units. \frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i\cdotN_{i-1}+D_{i-1}}{D_{i-1}\cdotF_i} RE: Testing the mathjax engine - Csaba Tizedes - 02-19-2017 03:40 PM (05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$ Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units. $$\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i · N_{i-1}+D_{i-1}}{D_{i-1} · F_i}$$ JÍHÁÁÁÁ! RE: Testing the mathjax engine - Luigi Vampa - 02-19-2017 08:14 PM (05-08-2016 08:56 PM)emece67 Wrote:  [•••] Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units. I might be wrong, I haven't been involved in thermal issues since my second course in the University. Kelvin degrees = Centigrade degrees + the absolute zero offset Offsets are canceled in the Delta, so that there would be no incoherence. The same Delta in Centigrade degrees would have the very same value. Fahrenheit degrees are another story, you know. RE: Testing the mathjax engine - Csaba Tizedes - 06-11-2017 10:56 AM (02-19-2017 03:40 PM)Csaba Tizedes Wrote:   (05-08-2016 08:56 PM)emece67 Wrote:  From $$Q = mC\Delta T$$ you will get $$\Delta T = {Q\over mC}$$, with $$C=4182 {J\over kg \Delta K}$$ being the specific heat of water at 20ºC. With this you get $\Delta T = {25000 J\over 2kg · 4182 {J\over kg \Delta K}} = 2.989 \Delta K$ Then the final temperature will be: $$T_0+\Delta T = 20 ºC + 2.989 \Delta K =$$ Inconsistent units. $$\frac{N_i}{D_i}=\frac{N_{i-1}}{D_{i-1}}+\frac{1}{F_i}=\frac{F_i · N_{i-1}+D_{i-1}}{D_{i-1} · F_i}$$ JÍHÁÁÁÁ! From these equations ($$\beta$$ in radian): $$R_2 · \beta=2 · r · \pi$$, and $$(R_2+s) · \beta=2 · R · \pi$$ we get: $$R_2=\frac{\frac{h}{cos \alpha}}{\frac{R}{R-h_x}-1}$$, and $$\beta=\frac{2 · (R-h_x) · \pi}{R_2}$$