Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡ HI ≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴
Due to Rectilinear propagation of light
- A shadow is formed
- Formation of Day and Night
- An Image in the pinhole camera is formed
∞║║ HOPE THIS HELPS PLEASE MARK AS BRAINLIEST║║∞
