A container of 5 L water is left out on a hot day. A temperature change of 2.5°C was noted. Estimate the heat transfer that took place. (Ignore all other heat loss)
A container of 5 L water is left out on a hot day. A temperature change of 2.5°C was noted. Estimate the heat transfer that took place. (Ignore all other heat loss)
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Transcribed Image Text:**Question:**
A container of 5 L water is left out on a hot day. A temperature change of 2.5°C was noted. Estimate the heat transfer that took place. (Ignore all other heat loss)
**Explanation for Educational Use:**
This problem involves calculating the heat transfer that occurs when a known volume of water undergoes a temperature change. To estimate the heat transfer, we can use the formula \( Q = mc\Delta T \), where:
- \( Q \) is the heat transfer (in joules),
- \( m \) is the mass of the water (in kilograms),
- \( c \) is the specific heat capacity of water (approximately \( 4,186 \, \text{J/kg}^\circ\text{C} \)),
- \( \Delta T \) is the change in temperature (in degrees Celsius).
Given:
- Volume of water = 5 L (which is equivalent to 5 kg, assuming the density of water is \( 1 \, \text{kg/L} \)),
- Temperature change (\( \Delta T \)) = 2.5°C.
Substitute these values into the equation to find the heat transfer.
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