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What is the identity of a 100.0 g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0 °C? Metal Specific Heat Capacity J/(g.°c) Aluminum 0.90 Copper 0.38 Iron 0.45 Magnesium 1.00 Silver 0.23 Tin 0.32 Titanium 0.52 A) Mg B) AI C) Ti D) Fe

Respuesta :

The metal is the Aluminium that has the 100.0 g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0 °C.

We have a mass (m) of 100. g of an unknown steel that, while absorbs 4680 J of warmth (Q) will increase in temperature via way of means of 52.0°C (AT). The unique warmness capability (c) of the steel is the quantity of warmth required to elevate the temperature of the unit mass via way of means of the unit of temperature.

We can calculate it the usage of the subsequent expression.

  • c = Q/(m*Delta*T) = 4680J 100.g*52.0^ C = 0.900J/g.°C
  • Aluminum is the steel whose unique warmness capability is 0.900 J/g. °C.
  • Aluminum is a steel that, in a 100. g pattern soaking up 4680 J of warmth will increase in temperature via way of means of 52.0 °C.

Read more about heat capability here: https://brainly.com/question/22563191

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