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How much would the temperature (in °C) of 259 g of water at 25°C increase if 36.0 kJ of heat were added?

User Ketan Modi
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Final answer:

The temperature of 259g of water would increase by approximately 33.5°C when 36.0 kJ of heat is added, calculated using the formula q = mcΔT with the specific heat capacity of water.

Step-by-step explanation:

To calculate the increase in temperature for a specific mass of water when a certain amount of heat is added, we use the formula q = mcΔT, where q is the heat added in Joules, m is the mass of the water in grams, c is the specific heat capacity (4.184 J/g°C for water), and ΔT is the change in temperature in °C.

Given that 36.0 kJ (or 36000 J) of heat is added to 259 g of water at 25°C, we can rearrange the formula to solve for ΔT: ΔT = q / (mc).

ΔT = (36000 J) / (259 g × 4.184 J/g°C)

ΔT ≈ 33.5°C (rounded to one decimal place)

Therefore, the temperature of the 259 g of water would increase by approximately 33.5°C when 36.0 kJ of heat is added.

User Breno
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