1 Answer | Add Yours
Small glass cup, 180 grams piece of ice taken from the fridge, thermometer, 10 V voltage DC voltage source, 10 ohm/20 W resistor, chronometer.
Place the ice into the cup and measure its temperature. Wire the resistor to the voltage source, fix the voltage generated to exactly 10 V and place the resistor on top of ice. Note the time shown by the clock.
Let about two hours for the ice to melt. Stop the power source, and measure again the temperature of the water. Note the time when the experiment has ended.
Data and Computations
Because the glass cup has a small mass and the specific heat capacity of the glass is small, the heat taken by the glass will be small enough compared to the heat taken bythe ice and the water after the ice melts.
Initial temperature of ice T =-7 degree C; final temperature of water T =5 degree C. Time indicated by clock t =111 minutes=6660 seconds
Heat delivered by resistor is
`Q =(U^2/R)*t = 100/10*6660 =66600 J`
Heat taken by the ice at -7 degree to melt and transform into water at 5 degree is
`Q =m*c_("ice")*7 +m*lambda_("ice") +m*c¨_("water")*5`
`66600 =180*7*2.11 +180*lambda_("ice") +180*5*4.18`
`lambda_("ice") = 334.33 J/g`
Per one mole of water (`M(H20) =2+16 =18 g` )
`lambda_("ice") = 334*18 =6017.94 J/("mole")`
We’ve answered 333,813 questions. We can answer yours, too.Ask a question