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When an amount of heat Q (in kcal) is added to a unit mass (kg) of a
substance, the temperature rises by an amount T (°C). The quantity dQ/dT, called
specific heat, is 0.18 (kcal/°C) for glass. If dQ/dt = 12.9 kcal/min for a 1 kg sample
of glass at 20.0 °C, find dT/dt for this same sample. (Note: T is for temperature, t is
for time.)

When an amount of heat Q in kcal is added to a unit mass kg of a substance the temperature rises by an amount T C The quantity dQdT called specific heat is 018 class=

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Answer:

The change in temperature per minute for the sample, dT/dt is 71.[tex]\overline {6}[/tex] °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

[tex]\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)[/tex]

[tex]\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)[/tex]

Therefore, we get;

[tex]\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}[/tex]

[tex]\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min[/tex]

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

[tex]\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min[/tex]

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