Question

40 POINTS!!! Ice at 0.0 degrees C is combined with 50.0g of water at 75.0 degrees C. Calculate the grams of ice present initially if the entire mixture comes to a final temperature of 25.0 degrees C after the ice melts. Specific heat of water is 4.18 J/goC. SHOW WORK!

Answer 1

mi = 31.28 g

Step-by-step explanation:

According to the law of conservation of energy:

`Heat\ Gain\ by\ Ice = Heat\ Lost\ by\ Water\\m_iL = m_wC\Delta T\\`

where,

mi = mass of ice = ?

L = Latent heat of fusion of ice = 334 J/g

mw = mass of water = 50 g

C = specific heat of water = 4.18 J/g.°C

ΔT = change in temperature of water = 75°C - 25°C = 50°C

Therefore,

`m_i(334\J/g) = (50\ g)(4.18\ J/g.^oC)(50^oC)`

mi = 31.28 g