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50-g of hot water at 65 degree C is poured into a cavity in a very large block of ice at 0 degrees C. The final temperature of the water in the cavity is then 0 degrees C. What is the mass of the ice that melts?
Solve using calories. Show equations, units, and work.

Respuesta :

Answer:

m = 40.88 g

Explanation:

Given data:

We know that Ice at zero degree celcius convert into liquid form by observing heat

latent heat of fusion = 79.5 cal/g

heat required or relaesed to convert 50 gram of water at 65 degree  c to zero degree celcius

let mass of ice is m

[tex]50 \times\ (1 cal/g) \times (65-0) = m(79.5) cal/g[/tex]

where 1 cal/g  os specific heat of water

solving for m

[tex]m = \frac{50\times 65}{79.5} g[/tex]

m = 40.88 g

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