Question

A 73.9 g silver block with a specific heat of 0.235 J/g °C, is submerged into 100.0 g of water at 24.80c, in an insulated container increases to 26.20c. Determine the initial temperature ofthe silver block and the water.

Answer 1

To find the initial temperatures of the silver block and water, we can use the concept of heat transfer and the equation Q = mcΔT.

Step-by-step explanation:

In this problem, we can use the concept of heat transfer and the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the silver block, the heat transferred can be calculated as follows:

Qs = mcΔTs = (73.9g)(0.235 J/g °C)(26.20 °C - Ts)

For the water, the heat transferred can be calculated as follows:

Qw = mcΔTw = (100.0g)(4.18 J/g °C)(26.20 °C - Tw)

Since the total heat transferred is equal to zero (insulated container), we can set Qs + Qw = 0 and solve for Ts and Tw to find the initial temperatures of the silver block and the water, respectively.