Question

When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

Answer 1

Answer:

The change in temperature per minute for the sample, dT/dt is 71.$\overline {6}$ °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;

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

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

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

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

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