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THE TEMPERATURE RAIED WHEN 1000CAL OF HEAT IS SUPPLIED TO 1GRAM OF WATER AT 25DEGREE CELSIUS

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Answer:

The temperature change (ΔT) is 1000 degrees Celsius.

Step-by-step explanation:

To calculate the temperature change when 1000 calories of heat is supplied to 1 gram of water at 25 degrees Celsius, we need to use the specific heat capacity of water. The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

For water, the specific heat capacity is approximately 1 calorie/gram°C, which means it takes 1 calorie of heat to raise the temperature of 1 gram of water by 1 degree Celsius.

Now, let's calculate the temperature change:

Calculate the heat energy required to raise the temperature of 1 gram of water by ΔT degrees Celsius:

Heat energy (Q) = mass (m) × specific heat capacity (c) × ΔT

Given:

mass (m) = 1 gram

specific heat capacity (c) = 1 calorie/gram°C

ΔT = unknown (final temperature - initial temperature)

Heat energy (Q) = 1000 calories

Rearrange the equation to solve for ΔT:

ΔT = Q / (m × c)

ΔT = 1000 calories / (1 gram × 1 calorie/gram°C)

ΔT = 1000 / 1 = 1000 degrees Celsius

The temperature change (ΔT) is 1000 degrees Celsius.

However, it's important to note that this result seems unusually high. The specific heat capacity of water is usually given in joules, not calories. If the specific heat capacity was given in joules/gram°C, the result would be in degrees Celsius. If you meant the specific heat capacity in joules, please let me know so I can recalculate the answer accordingly.

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