# Answer to Question #23441 in Inorganic Chemistry for riley

Question #23441

If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?

Expert's answer

1 cal = 4.1868 J

1 J = 0.238846 cal

15000 Calories (IT) = 62802 Joules

Heat capacity (usually denoted by a capital C, often with subscripts), or thermal capacity, is the measurable physical quantity that shows the amount of heat required to change the temperature of a substance by a given amount. In the International System of Units (SI), heat capacity is expressed in units of joule(s) (J) per kelvin (K) (or Celsius)

The specific heat of water is C = 4184 J/kg*°C, which is read 4184 Joules per kilogram degree Celsius. You have 2 L or 2 kg of water, so you need 4184*2 = 8368 J to heat this volume of water to change the temperature of one degree Celsius.

Temperature when you finished would be 25 + 62802/4184 = 40 C

1 J = 0.238846 cal

15000 Calories (IT) = 62802 Joules

Heat capacity (usually denoted by a capital C, often with subscripts), or thermal capacity, is the measurable physical quantity that shows the amount of heat required to change the temperature of a substance by a given amount. In the International System of Units (SI), heat capacity is expressed in units of joule(s) (J) per kelvin (K) (or Celsius)

The specific heat of water is C = 4184 J/kg*°C, which is read 4184 Joules per kilogram degree Celsius. You have 2 L or 2 kg of water, so you need 4184*2 = 8368 J to heat this volume of water to change the temperature of one degree Celsius.

Temperature when you finished would be 25 + 62802/4184 = 40 C

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