Question

Consider a 100 gm of copper taken at 60°C is cooled to 40°C . How much heat does it release ? [Specific heat of copper = 0.09 cal/g/°C]​

Answer 1

Answer:

Heat given by water to reach 5

o

C+ Heat given by copper vessel to reach 5

o

C= Heat taken by ice to melt at 0

o

C+ Heat taken by melted ice to reach 5

$o</p><p> C</p><p></p><p>m </p><p>w$

$c </p><p>w</p><p>$

$Δt+m </p><p>c</p><p> </p><p> c </p><p>c$

Δt=m

i

L+m

i

c

w

δt

where Δt= change in temperature of water and copper vessel

m

w

= mass of water = 150g

m

c

= mass of copper vessel = 100g

m

i

= mass of ice

c

c

= specific heat capacity of copper

c

w

= specific heat capacity of water

L = specific latent heat of fusion of ice

δt = change in temperature of ice

Therefore, [150×4.2×(50−5)]+[100×0.4×(50−5)]=(m×336)+[m×4.2×(5−0)]

(150×4.2×45)+(100×0.4×45)=(m×336)+(m×4.2×m)

28350+1800=336m+21m

357m=30150

m=

357

30150

=84.45g of ice

Hence, 84.45g of ice is needed to cool it to 5°C