A small 59 g water sample A is heated for 259 seconds by a 22.1 Watt heater resu
ID: 1495265 • Letter: A
Question
A small 59 g water sample A is heated for 259 seconds by a 22.1 Watt heater resulting in an increase temperature DeltaTA. The heated small water sample A is added to a larger tank of water, sample B. but. during the pouring process, water sample A cools down by 10% (a 10% decrease in the^TA). The larger tank of water sample B was originally 346 g of water, before sample A was added. What is the final temperature change ATa + g of the combined samples? Units are in degrees Celsius; do not include units in your answer. Use Cp = 4.1813 JfgdegreeC. anc give your answer to two decimal places.Explanation / Answer
sample 1
m1 = 59g
Heat added = power *time = 22.1*259 = 5723.9 J
H = mCdT
5723.9 = 59*4.1813*dTa
dTa = 23.20217
after 10 p.c. loss of temp, dTa = 20.88195
After adding to next sample
m1CdTa = (m1+m2)CdTa_b
59*20.88195 = (59+346)dTa_b
dTa-b = 3.0420