Imagine that you have two cups, one containing 1550 g of water and the other one
ID: 1475799 • Letter: I
Question
Imagine that you have two cups, one containing 1550 g of water and the other one containing some amount of an unknown liquid. The specific heat of water is 4.19 J/g·°C, and the specific heat of the unknown liquid is 6.81 J/g·°C. You use two identical immersion heaters to heat the water and the unknown liquid simultaneously. Assume that all the heat from the heater is used to heat the content of the cups.
(a) You started heating the water and the unknown liquid at the same time. If the temperature of the water was increased from 18.4 °C to 37.1 °C while the temperature of the unknown liquid is increased from 18.4 to 25.5 °C, what was the mass of the unknown liquid in the cup? ____grams
(b) If you used 184 W heaters to heat the liquids in part (a), how long did you heat the liquids? ____minutes
Explanation / Answer
a)
let m is the mass of umknown liquid.
here
heat obsorbed by water = heat obsorbed by unknown liquid
1550*4.19*(37.1-18.4) = m*6.81*(25.5 - 18.4)
m = 1550*4.19*(37.1-18.4)/(6.81*(25.5 - 18.4))
= 2512 grams <<<<<<<<<-------------------------------Answer
b) we know, Power = Energy delivered/time
==> time = Energy deliverd/power of heater
= 1550*4.19*(37.1-18.4)/184
= 660 s
= 11 minutes <<<<<<<<<-------------------------------Answer