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A sample of metal is heated and added to a calorimeter with 100. g of water at 25°C. The temperature of water changed a total of 15°C. How much heat is gained by the water? (cwater = 4.18 J / [ g . ° C ] )

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The heat gained by the water can be calculated using the formula:

\[ q = m \cdot c \cdot \Delta T \]

where:
- \( q \) is the heat gained,
- \( m \) is the mass of water,
- \( c \) is the specific heat capacity of water,
- \( \Delta T \) is the change in temperature.

In this case:
- \( m = 100 \, \text{g} \),
- \( c = 4.18 \, \text{J/g}^\circ\text{C} \),
- \( \Delta T = 15^\circ\text{C} \).

Substitute these values into the formula:

\[ q = 100 \, \text{g} \cdot 4.18 \, \text{J/g}^\circ\text{C} \cdot 15^\circ\text{C} \]

\[ q = 100 \, \text{g} \cdot 4.18 \, \text{J/g}^\circ\text{C} \cdot 15^\circ\text{C} \]

\[ q = 6270 \, \text{J} \]

Therefore, the heat gained by the water is \(6270 \, \text{J}\).
User Roman Pustylnikov
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