two identical calorimeter A and B contain equal quantity of water at 20 degree Celsius 5 gram piece of metal x of specific heat 0.2 calorie gram is dropped into A and 5 gram piece of metal Y into B the equilibrium temperature in A is 22 degree Celsius and in B is 23 degree Celsius the initial temperature of both the metal is 40 degree Celsius find the specific heat of the metal Y in calorie gram
Answers
Answer:
As we know the heat lost by meta is equal to the gain of the water i.e
Heat lost by metal= heat gain by water
therfore,
mxcx(Tin-Teq)= mwcw(Teq- Tin).............(1)
mycy(Tin-Teq)= mwcw(Teq- Tin).............(2)
according to the problem:
mxcx(40-22)= mwcw(22-20)................(3)
mycy(40-23)=mwcw(23-20).................(4)
Dividing equation 3 and 4 we get
cx/cy 18/17=2/3
cy= 27/17* 0.2 cal/g
cy= 0.317cal/g
Hence the specific heat of the metal Y is 0.317cal/g
Answer:
Explanation:
"It is known to us that the heat lost by the metal is equal to the heat gain by the water, that is,
Heat lost by metal= heat gain by water
therfore,
mxcx(Tin-Teq)= mwcw(Teq- Tin).............(1)
mycy(Tin-Teq)= mwcw(Teq- Tin).............(2)
according to the problem:
mxcx(40-22)= mwcw(22-20)................(3)
mycy(40-23)=mwcw(23-20).................(4)
Dividing equation 3 and 4 we get
cx/cy 18/17=2/3
cy= 27/17* 0.2 cal/g
cy= 0.317cal/g
Hence the specific heat of the metal Y is 0.317cal/g"