PYSICS HELP polz....................vry difficult question..dr zakir nike couldnt even solve it...?

Question

A bath contains 100kg of water at60 dergee C. Holad and cold taps are then turned on to deliever 20kg per minute each at temperatures of 70 degree C and 10 degree C.Hw long will it be before the temperature in the bath has dropped to 45 degree C. ??????assume complete mixing of water n neglect heat losses.??????????????????

Answer 1

Beats me and everyone else around here. :D Ask this question in a thermodinamics forum.
Good luck.

Answer 2

Adding 20kg per minute each at temperatures of 70 degree C and 10 degree C is the same as adding 40 kg per min. at the average temperature:
(70 + 10)/2 = 40 degree C.
m1Δt1 = m2Δt2
100(60 - 45) = x(45 - 40)
1500 = x(5)
x = 300kg
300kg/(40kg/min) = 7.5 min

Answer 3

this is not difficlt ...................
assuming tht the heat gained is equal to the heat lost
< gained by cold water n lost by hot water>

heat gained= mct
m= mass
c= specfic heat capacity of water
t= temperature in Kelvins

hot water in bath
mass= 100
c= 4120
t= 273 + 60= 333
hot water tht was poured latr
mass= x
c= 4120
t= 343
so frst of all we need to determine this mass......... bcz the 2 types of hot waters establish an equilibrium frst n then the equilibrium btwn .,......... hot n clold

heat gaind = heat lost
100* 4120* 333= x* 4120* 343
x= 100* 333/343
x= 97.08 kg

now we dnt know the temperature here
of the total hot water

i will solve the next bit latr........ im jst a bit confused here


but yh a lovely question

n yh if u cn wrk out the temperature of the hot water........ actually yh
u cn
wrk out the total energy in the total hot water seperately
like eneergy in bath water
e= 100* 4120* 333
e= 137196000 J

energy in other hot water
e= mct
e= 97.08* 4120*343
e= 137189572.8 J

total energy of the hot waters
add both of them
= 274385572.8 J

now wrk out the temperute
e= mct
274385572.8 = (100+ 97.08)* 4120* t

wrk out the tempertue
t= 337.9 k

now mixing of hot n cold water
[197.08 * 4120* 337.9] + [x < cold water> * 4120* <273+10>]all this is the water in the bath now by adding all the water
this equals to water a 45 degree c
<197.08 + x>*4120* <273+45>
(197.08+x) * 4120 *218

now heat in this cold water is equal to the heat in the total hot water

so equal both of them
<197.08 + x> * 4120* 218 = 274385572.8 + (x* 4120 * 283)

177009372.8 + 898160 x = 274385572.8 + 1165960x

177009372.8 - 274385572.8 = 1165960x- 898160x
267800x = -97376200
now i think i hv made a mistajke somewher in calculation bcz i got a negative answer
n mass cnt b negative

but this is hw it is gng to wrk

nw wen u get the mass of water
then divide it by 20, u will get the no. of minutes for which the water needs to b on
n do the same for hot water as well as we calculated earlier
n yh please do email me the ansdwer

tk cr m8
i hope u get it rite