If you wish to warm 110kg of water by 18 degrees C for your bath how much heat is required?
Heat= mass*specificheat*TempChange
Put the mass in kg, choose the specific heat constant in units of J/(kg-C), and use the 18C as the temp change.
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To calculate the amount of heat required, we will use the formula:
Heat = Mass * Specific Heat * Temperature Change
Given:
Mass of water (m) = 110 kg
Specific heat of water (c) = 4186 J/(kg·°C)
Temperature change (ΔT) = 18 °C
Substituting these values into the formula, we get:
Heat = 110 kg * 4186 J/(kg·°C) * 18 °C
Now, let's do the calculation:
Heat = 110 kg * 4186 J/(kg·°C) * 18 °C
= 838,680 J/°C * kg * 18 °C
Calculating this further, we have:
Heat = 838,680 J/°C * kg * 18 °C
= 15,096,240 J
Therefore, you would need approximately 15,096,240 Joules of heat to warm 110 kg of water by 18 degrees Celsius.