Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate the specific heat of the metal (J/goC). Assume that all of the heat lost by the metal is transferred to the water and no heat is lost to the surroundings.

Sagot :

Lanuel

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

  • Initial temperature of water = 25.0°C
  • Final temperature of water = 29.0°C
  • Mass of water = 400.0 g
  • Mass of metal = 30.0 g
  • Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

[tex]Q_{metal} = Q_{water}[/tex]

Mathematically, heat capacity or quantity of heat is given by the formula;

[tex]Q = mc\theta[/tex]

Where:

  • Q is the heat capacity or quantity of heat.
  • m is the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

Substituting the values into the formula, we have:

[tex]30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}[/tex]

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

Read more: https://brainly.com/question/18691577

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.