Solution:
According to the question Mass of ice (m1) is 17 g
Mass of water m2 is 40 g
Change in temperature is given by
34 – 0 = 34 K
We know that the specific heat capacity of water is 4.2 J g-1 K-1
Heat energy gained by ice (latent heat of ice), Q = heat energy released by water, assuming no heat loss. Using the below expression for heat energy,
Q = m × c × (Change in temperature)
Upon substituting values, we get => Q = 40 × 34 × 4.2
Q = 5712 J
Expression for the specific latent heat of ice is
L = Q / m = 5712 / 17
Therefore, L = 336 J g-1
Assumption : There is no loss of energy to the surroundingd.