The specific heat of water is approximately 4.186 joules per gram per degree Celsius (J/g°C), which can also be expressed as 4186 joules per kilogram per degree Celsius (J/kg°C).
To calculate the heat contained in 100 kg of water at 60.0 °C, you can use the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( Q \) is the heat (in joules),
- \( m \) is the mass of the water (in kilograms),
- \( c \) is the specific heat capacity (in J/kg°C),
- \( \Delta T \) is the temperature change (in °C).
In your case, if you want to calculate the heat \( Q \) contained in water at a specific temperature (without a reference temperature for the change), then Q can be treated just as the heat at that temperature (assuming the reference for this calculation is 0°C):
- \( m = 100 \) kg (mass of the water),
- \( c = 4186 \) J/kg°C (specific heat of water),
- The temperature at which you're measuring is \( T = 60.0 \) °C, but without a starting temperature to define \( \Delta T \) for heat content evaluation, we can consider that the heat content can regard it as it is at the given temperature.
However, if you want to find the total energy in this mass of water, the specific heat does not play a role without an additional reference. So typically, heat is related to a temperature difference from a point of reference.
If we assume that the reference point is 0°C and thus \( \Delta T = 60.0 - 0 = 60.0 °C \), then we can calculate the heat as follows:
\[ Q = 100 , \text{kg} \cdot 4186 , \text{J/kg°C} \cdot 60.0 , \text{°C} \]
Calculating this, we have:
\[ Q = 100 \cdot 4186 \cdot 60.0 = 25116000 , \text{J} \]
So, the heat contained in 100 kg of water at 60.0 °C is 25,116,000 J (or 25.1 MJ).
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