Calculate the heat required to warm 16.9 g of water from 87.4 C to 100.0 C.
The correct answer and explanation is:
To calculate the heat required to warm water, we use the specific heat capacity formula: Q=mcΔTQ = mc\Delta T
Where:
- QQ is the heat energy (in joules, J),
- mm is the mass of the substance (in grams, g),
- cc is the specific heat capacity of the substance (for water, it is 4.18 J/g°C4.18 \, \text{J/g°C}),
- ΔT\Delta T is the change in temperature (in °C).
Step 1: Identify the given values
- Mass m=16.9 gm = 16.9 \, \text{g},
- Specific heat capacity c=4.18 J/g°Cc = 4.18 \, \text{J/g°C},
- Initial temperature T1=87.4°CT_1 = 87.4°C,
- Final temperature T2=100.0°CT_2 = 100.0°C.
Step 2: Calculate the temperature change
The temperature change ΔT\Delta T is: ΔT=T2−T1=100.0°C−87.4°C=12.6°C\Delta T = T_2 – T_1 = 100.0°C – 87.4°C = 12.6°C
Step 3: Apply the formula
Now, substitute the values into the formula: Q=(16.9 g)×(4.18 J/g°C)×(12.6°C)Q = (16.9 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (12.6°C) Q=16.9×4.18×12.6=876.7 JQ = 16.9 \times 4.18 \times 12.6 = 876.7 \, \text{J}
Final Answer:
The heat required to warm 16.9 g of water from 87.4°C to 100.0°C is 876.7 J.
Explanation:
When you heat a substance, the amount of energy required depends on the mass, the specific heat capacity, and the temperature change. The specific heat capacity of water is relatively high, meaning it requires a significant amount of energy to raise its temperature. In this case, a relatively small mass of water (16.9 g) requires 876.7 joules of energy to be warmed by 12.6°C. This calculation assumes there is no phase change (such as water turning into steam), which is the case here since the temperature does not exceed 100°C.