| Common units | |
| Calorie/Gram °C | cal/g∙°C |
| Joule/Gram °C | J/g∙°C |
| Joule/Kilogram K | J/kg∙K |
| Joule/Kilogram °C | J/kg∙°C |
| Kilocalorie/Kilogram °C | kcal/kg∙°C |
| Kilojoule/Kilogram °C | kJ/kg∙°C |
| Kilojoule/Kilogram K | kJ/kg∙K |
| Other units | |
| BTU/Pound °C | BTU/lb∙°C |
| BTU/Pound °F | BTU/lb∙°F |
| BTU/Pound °R | BTU/lb∙°R |
| CHU/Pound °C | CHU/lb∙°C |
| Common units | |
| Calorie/Gram °C | cal/g∙°C |
| Joule/Gram °C | J/g∙°C |
| Joule/Kilogram K | J/kg∙K |
| Joule/Kilogram °C | J/kg∙°C |
| Kilocalorie/Kilogram °C | kcal/kg∙°C |
| Kilojoule/Kilogram °C | kJ/kg∙°C |
| Kilojoule/Kilogram K | kJ/kg∙K |
| Other units | |
| BTU/Pound °C | BTU/lb∙°C |
| BTU/Pound °F | BTU/lb∙°F |
| BTU/Pound °R | BTU/lb∙°R |
| CHU/Pound °C | CHU/lb∙°C |
What is the final temperature of 420.1 g of water (specific heat = 4.18 J/g･°C) at 24.2°C that absorbed 950. J of heat?
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Q: A quantity of 505 J of heat are added to 11.3 g of water at 28.3 °C. What is the final temperature…
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Q: A quantity of 505 J of heat are added to 11.3 g of water at 28.3 °C. What is the final temperature…
A: According to the answering guidelines, I'm posting the solution for the first question. Kindly post…
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