# How many joules are required to heat 250 grams of liquid water from 0°C to 100°C?

The crucial formula is

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The specific heat capacity of water is 4.18 J/g°C. The formula to calculate the energy required to heat a substance is Q = mcΔT, where Q is the energy in joules, m is the mass in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature in degrees Celsius. Substituting the values, we get: Q = (250 g) * (4.18 J/g°C) * (100°C - 0°C) = 104500 J. Therefore, 104500 joules are required to heat 250 grams of liquid water from 0°C to 100°C.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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