80gm of water at 30°C is poured on a large block of ice at 0°C. The mass of the ice that melts is?
Answers
Answered by
319
A/C to thermodynamics ,
Heat flow from higer temperature body to lower temperature body
Heat loss by water = Heat gain by ice
Given, mass of water = 80gm
specific heat of water = 1 Cal/gm/°C
∴ Heat loss by water = MS∆T
= 80 × 1 × (30 -0)
= 80 × 1 × 30
= 2400cal
Heat gain by ice = MLf
Lf is latent heat of fusion = 80Cal/gm
M is the mass of melted ice
Now, 2400 = 80M
M = 30gm
Hence , 30gm of ice is melted .
Heat flow from higer temperature body to lower temperature body
Heat loss by water = Heat gain by ice
Given, mass of water = 80gm
specific heat of water = 1 Cal/gm/°C
∴ Heat loss by water = MS∆T
= 80 × 1 × (30 -0)
= 80 × 1 × 30
= 2400cal
Heat gain by ice = MLf
Lf is latent heat of fusion = 80Cal/gm
M is the mass of melted ice
Now, 2400 = 80M
M = 30gm
Hence , 30gm of ice is melted .
Answered by
20
Answer:
80*30=2400/80=30g
Explanation:
Heat lost =heat gained
Similar questions