Answer to Question #166981 in Molecular Physics | Thermodynamics for don

Question

A cup is made of an experimental material that can hold hot liquids without significantly

increasing its own temperature. The 0.80 kg cup has an initial temperature of 37 degrees C

when is submerge in 1.54 kg of water with an initial temperature of 20 degrees C. What is the

specific heat capacity if the final temperature is 27 degrees C?A cup is made of an experimental material that can hold hot liquids without significantly

increasing its own temperature. The 0.80 kg cup has an initial temperature of 37 degrees C

when is submerge in 1.54 kg of water with an initial temperature of 20 degrees C. What is the

specific heat capacity if the final temperature is 27 degrees C?

Answer 1

$Q_1=-Q_2,$

$Q_1=c_1m_1(t-t_1),$

$Q_2=c_2m_2(t-t_2),$

$c_1=c_2\cdot \frac{m_2}{m_1}\cdot\frac{t-t_2}{t_1-t},$

$c_1=4184 \cdot \frac{1.54}{0.8}\cdot\frac{27-20}{37-27}=5.636~\frac{kJ}{kg\cdot °C}$