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Calculate the heat required to raise 31.3 g of iron (0.108 cal/gx°C) from 22.0 °C to 69.0 °C.
Express your answer with the appropriate units.
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Answer :

GinnyAnswer
To calculate the heat required to raise the temperature of 31.3 g of iron from 22.0 °C to 69.0 °C, we can use the specific heat formula:

[tex]\[ Q = mc\Delta T \][/tex]

where:
- [tex]\( Q \)[/tex] is the heat energy in calories (cal)
- [tex]\( m \)[/tex] is the mass in grams (g)
- [tex]\( c \)[/tex] is the specific heat capacity in calories per gram per degree Celsius (cal/g°C)
- [tex]\( \Delta T \)[/tex] is the change in temperature in degrees Celsius (°C)

Here are the given values:
- Mass ([tex]\( m \)[/tex]) = 31.3 g
- Specific heat ([tex]\( c \)[/tex]) = 0.108 cal/g°C
- Initial temperature [tex]\( (T_i) \)[/tex] = 22.0 °C
- Final temperature [tex]\( (T_f) \)[/tex] = 69.0 °C

First, we need to determine the temperature change ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = T_f - T_i \][/tex]
[tex]\[ \Delta T = 69.0 °C - 22.0 °C \][/tex]
[tex]\[ \Delta T = 47.0 °C \][/tex]

Next, we calculate the heat required ([tex]\( Q \)[/tex]):
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = (31.3 \text{ g}) \times (0.108 \text{ cal/g*°C}) \times (47.0 °C) \][/tex]

Multiplying these values together gives:
[tex]\[ Q = 158.8788 \text{ cal} \][/tex]

So, the heat required is:

[tex]\[ \boxed{158.9 \text{ cal}} \][/tex]

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