Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific latent heat of ice = 336 J g^-1, and specific heat capacity of water = 4.2 J g^-1 K^-1.
Calorimetry
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Question

Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific latent heat of ice = 336 J g^-1, and specific heat capacity of water = 4.2 J g^-1 K^-1.

Calorimetry

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KnowledgeBoat · Accepted Answer

Given, mass of ice = 10 g mass of water = 10 g Specific heat capacity of ice = 2.1 J g^-1 K^-1 specific latent heat of ice = 336 J g^-1, specific heat capacity of water = 4.2 J g^-1 K^-1 Let the final temperature be t° C. Heat energy taken by ice at – 10° C to raise it's temperature to 0° C (Q₁) = m x c x change in temperature = 10 x 2.1 x [0 - (-10)] = 10 x 2.1 x 10 = 210 J Hence, Q₁ = 210 J Heat energy taken by ice at 0° C to convert into water at 0° C (Q₂) = m x L_ice = 10 × 336 = 3360 J Hence, Q₂ = 3360 J Heat energy taken by water at 0° C to raise it's temperature to t° C (Q₃) = m x c x change in temperature = 10 × 4.2 × (t – 0) = 10 × 4.2 × t = 42 t Hence, Q₃ = 42 t Heat energy released by water at 10° C to lower it's temperature to t° C (Q₄) = m x c_water x change in temperature = 10 × 4.2 × (10 – t) = 42 × (10 – t) = 420 – 42t Hence, Q₄ = 420 – 42t If there is no loss of heat, Heat energy gained = Heat energy lost This cannot be true because water cannot exist at -37.5° C. Hence, we can say that the whole of the ice did not melt. Let amount of ice which melts = m g Final temperature of the mixture = 0° C Heat energy gained by ice at -10° C to raise it's temperature to 0° C = m x c_ice x change in temperature = 10 × 2.1 x [0 - (-10)] = 10 × 2.1 x 10 = 210 J Heat energy gained by m gm of ice at 0° C to change into water at 0° C = m × L_ice = m x 336 = 336m J Heat energy released by 10 g of water at 10° C to lower it's temperature to 0° C = m x c_water x change in temperature = 10 × 4.2 × (10 – 0) = 10 × 4.2 × 10 = 420 J If there is no loss of heat, Heat energy gained = Heat energy lost Hence, 0.625 g of ice will melt and temperature will remain at 0° C

Physics

Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific latent heat of ice = 336 J g^-1, and specific heat capacity of water = 4.2 J g^-1 K^-1.

Calorimetry

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In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34° C to it's freezing temperature. The specific heat capacity of water is 4.2 J g^-1 K^-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of ice = 336000 J kg^-1.

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Physics

Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g-1 K-1, specific latent heat of ice = 336 J g-1, and specific heat capacity of water = 4.2 J g-1 K-1.

Calorimetry

Related Questions

In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34° C to it's freezing temperature. The specific heat capacity of water is 4.2 J g-1 K-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of ice = 336000 J kg-1.

A piece of ice of mass 40 g is added to 200 g of water at 50° C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice = 336 x 103 J kg-1.

Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1 °C-1, Latent heat capacity of ice = 330 J g-1.

Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific latent heat of ice = 336 J g^-1, and specific heat capacity of water = 4.2 J g^-1 K^-1.

In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34° C to it's freezing temperature. The specific heat capacity of water is 4.2 J g^-1 K^-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of ice = 336000 J kg^-1.

A piece of ice of mass 40 g is added to 200 g of water at 50° C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of fusion of ice = 336 x 10³ J kg^-1.

Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g^-1 °C^-1, Specific heat capacity of water = 4.2 J g^-1 °C^-1, Latent heat capacity of ice = 330 J g^-1.