Question

Ex. 2 A student places 50.0mt of गiquid water at 21.00 ∘ C into a coffee cup calorin re of She places a sample of gold at 100.00 ∘ C into the calorimeter. The final temperature of the water is 21.33 ∘ C. The density of water is d=1.00 g/mL. Calculate the quantity of thermal energy absorbed by the water in the calorimeter. V H20 ​ =50.0 mLA:q=MaC⋅ AT

Answer 1

To solve this problem, we will be using an equation that relates the heat absorbed by a material, its mass, its specific heat capacity, and the change in temperature: q = mcΔT, where:

- q is the quantity of heat absorbed by the material,
- m is the mass of the material,
- c is the specific heat capacity of the material, and
- ΔT is the change in temperature.

In this problem, the mass of water is 50.0 g, the specific heat capacity of water is 4.18 J/gC, the initial temperature of the water is 21.0 °C, and the final temperature is 21.33 °C.

First, we need to calculate the change in temperature for the water. This can be found by subtracting the initial temperature from the final temperature:

ΔT (temperature change) = T_final - T_initial
= 21.33 °C - 21.00 °C
= 0.33 °C

Now that we know the change in temperature, we can use this information to find out how much heat the water absorbed.

Using the equation for heat absorption:

q = mcΔT
= (50.0 g) * (4.18 J/g°C) * (0.33 °C)
= 68.97 J

The water absorbed approximately 68.97 joules of thermal energy.