Question
A thermometer of mass 0.0550 kg and of specific heat 0.837 kJ/kg•K reads 15.0°C. It is then completely immersed in 0.300 kg of water, and it comes to the same final temperature as the water. If the thermometer reads 44.0 °C, what was the temperature of the water before the insertion of the thermometer?
Answers
The temperature of the water will drop as heat is added to the thermometer as it reaches an equilibrium temperature.
The specific heat of the water is 4.18 kJ/Kg*K
heat gained by thermometer
= heat lost by water
0.055*(0.837)(44 - 15) = 0.300*(4.18)(T - 44)
1.335 = 1.254 (T-44)
T - 44 = 1.08
T = 45.08 C is the initial water temperature
The specific heat of the water is 4.18 kJ/Kg*K
heat gained by thermometer
= heat lost by water
0.055*(0.837)(44 - 15) = 0.300*(4.18)(T - 44)
1.335 = 1.254 (T-44)
T - 44 = 1.08
T = 45.08 C is the initial water temperature