Answer :
Final answer:
If 35.5 kJ of heat were added to 256 g of water initially at 25 °C, its temperature would increase by approximately 33.5°C by using the formula ΔT = q / (m * c) and substituting the given values into it.
Explanation:
The subject here is about the thermal energy transfer involved in heating water, a concept in Physics. To calculate the increase in temperature when a certain amount of heat is added to a given mass of a substance, we use the formula: q = mcΔT, where q is the heat added, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we would have to rearrange the formula to solve for ΔT, which gives us ΔT = q / (mc). The specific heat capacity (c) of water is 4.184 J/g°C, the amount of heat added (q) is 35500 J (since 1 kJ = 1000 J), and the mass (m) of the water is 256 g.
So, substituting these values into the equation, we get:
ΔT = 35500 J / (256 g * 4.184 J/g°C) = 33.5 °C
Therefore, if 35.5 kJ of heat were added to 256 g of water initially at 25 °C, its temperature would increase by approximately 33.5°C.
Learn more about Thermal energy transfer here:
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