00:01
Here in part a, water, temperature of water is equal to 35 degree fahrenheit, that is 5 by 9 into 35 minus 32 degrees celsius, that is 1 .67 plus 273 .15 kelvin, that is equal to 274 .8 2 .8.
00:26
Now body temperature of body that is equals to 98 .6 degree fahrenheit now from here we have 5 by 9 multiply with 98 .6 minus 32 .0 degrees celsius here we get 37 .0 plus 27 .15 kelvin we get 310 .15 kelvin we get 310 .15 kelvin so now now delta s of cold water, delta as of cold water we have d change in integration of dco by dt is equal to mvc multiply integration of water body which is dt by t t t that is equals to mwc multiply natural logo temperature of body divided by temperature of water.
01:23
Here we get delta s of body is equal to minus modular source of of q divided by temperature of body is equal to minus m of water c t of body minus t of body minus t of water divided by a t of body that is delta s of system is equal to more or less of minus t of body that is equal to minus m of w c temperature of body minus temperature of body delta s of cold water plus delta s of body we get here that is equal to 0 .454 kilogram multiply with 4186 jule per kilogram per degree celsius 6 cool per kilogram to calvin which is equal multiply with multiply with 310 .5 minus 274 .82 divided by 310 .15, that is equals to 134 .0 .0 .0 .0.
02:42
Now, in part b, we have q hought minus q cold.
02:49
Here, hode minus q cold.
02:54
So now, here we have to put the value that is, mwc into tf minus t water that is equals to m minus math into c tf minus t body...