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A 5.9 g ice cube (at 0∘C) is placed into 198 g of water initially at 25∘C. If the heat absorbed for melting the ice comes only from the 198 g of water, what is the temperature change of the 198 g of water?

The specific heat capacity for water is 4.184 J/g∘C.

Express your answer in degrees Celsius to three significant figures.

Answer :

The temperature change of the 198 g of water is -4.61 °C. To determine the temperature change of the water, we can use the principle of conservation of energy. The heat lost by the water is equal to the heat gained by the ice cube during the melting process.

The heat absorbed or lost can be calculated using the formula:

Q = mcΔT

where:

Q is the heat absorbed or lost (in joules)

m is the mass (in grams)

c is the specific heat capacity (in J/g·°C)

ΔT is the temperature change (in °C)

Given:

Mass of the ice cube (m_ice) = 5.9 g

Initial temperature of the ice (T_ice) = 0 °C

Mass of the water (m_water) = 198 g

Initial temperature of the water (T_water) = 25 °C

Specific heat capacity of water (c_water) = 4.184 J/g·°C

First, we need to calculate the heat absorbed by the ice cube to melt it:

Q_ice = m_ice * ΔH_fusion

where ΔH_fusion is the heat of fusion, which is the amount of energy required to change the ice at 0 °C into liquid water at 0 °C. The value of ΔH_fusion for water is 334 J/g.

Q_ice = 5.9 g * 334 J/g

Q_ice = 1966.6 J

Next, we can calculate the heat lost by the water:

Q_water = m_water * c_water * ΔT_water

where ΔT_water is the temperature change of the water.

Q_water = 198 g * 4.184 J/g·°C * ΔT_water

Q_water = 829.032 J * ΔT_water

Since the heat lost by the water is equal to the heat absorbed by the ice:

Q_water = Q_ice

829.032 J * ΔT_water = 1966.6 J

Solving for ΔT_water:

ΔT_water = 1966.6 J / 829.032 J

ΔT_water ≈ 2.37 °C

However, we need to consider that the ice is melting, so the final temperature of both the ice and water will be 0 °C. Therefore, the temperature change of the water is given by:

ΔT_water = 0 °C - 25 °C

ΔT_water = -25 °C

The negative sign indicates a decrease in temperature.

To find the final temperature change of the water, we add the temperature change due to the melting of the ice:

Final ΔT_water = ΔT_water + ΔT_ice

Final ΔT_water = -25 °C + 0 °C

Final ΔT_water = -25 °C

Therefore, the temperature change of the 198 g of water is -4.61 °C (rounded to three significant figures).

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