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How many calories are needed to heat 33.1 grams of iron from 17 deg C to 48 deg C if the specific heat and of iron is 0.108 cal per degree C per gram?
Wiki User
∙ 6y ago
Q=m⋅c⋅ΔT
where:
�
m is the mass of the substance (in grams),
�
c is the specific heat capacity of the substance (in cal/°C·g),
Δ
�
ΔT is the change in temperature (in °C).
In this case, for iron:
�
=
33.1
m=33.1 grams,
�
=
0.108
c=0.108 cal/°C·g,
Δ
�
=
48
−
17
=
31
ΔT=48−17=31 °C.
Now, plug these values into the formula:
�
=
33.1
g
⋅
0.108
cal/°C
\cdotp
g
⋅
31
°C
Q=33.1g⋅0.108cal/°C\cdotpg⋅31°C
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