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Use Newtons Law of Cooling to determine the current temperature of a body in air temperature of 25°C if 45 minutes ago the body was 160°C and 10 minutes ago it was 90°C.

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In the table below for E106, which of the following is the amount of heat lost by the metal specimen when dropped in the calorimeter with water? TRIAL 1 Determining the Specific Heat of Metal Specimen Metal Specimen 1 Mass of calorimeter, me Mass of calorimeter with water Mass of water, mi Mass of metal specimen, mm Initial temperature of metal, tom Initial temperature of calorimeter, toc Initial temperature of water, tow Final temperature of the mixture, tmix Specific Heat of Water cw Specific Heat of calorimeter ce Experimental specific heat of metal, cm Actual specific heat of metal, cm Percentage Error O 699 cal O 212 cal O 675 cal O 488 cal 150.0 g 225.0 g 8 100.0 g 100°C 20.0°C 20.0°C 26.5°C 1.0000 çal/gC 0.2174 cal/gC cal/gC 0.0917 cal/gC %
The amount of heat is often measured in calories. If I add 3 calorie of heat energy to 1 gram of water, the temperature of the water will decrease by 1° stay the same, since water has a very high specific increase by 3° increase by 3° decrease by 2°C
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Use Newton's Law of Cooling