Question

A0.0025 -kg block of an unknown substance is at temperture of 82 C. it is plased is a calorimeter with 0.025 kg of water at 22C. the system reaches an equilibirium temperature of 27Cwhat is the unknown substance.??

Answer 1

Answer on Question #72691, Physics / Other

A 0.0025 kg block of an unknown substance is at temperature of $82{}^{\circ}\mathrm{C}$. It is placed in a calorimeter with 0.025 kg of water at $22{}^{\circ}\mathrm{C}$. The system reaches an equilibrium temperature of $27{}^{\circ}\mathrm{C}$. What is the unknown substance?

Solution:

The relationship between heat and temperature change is usually expressed in the form shown below where $c$ is the specific heat.

$$Q = c m \Delta T$$

The temperature change is symbolized by $\Delta T$ where

$$\Delta T = \text{Final temperature} - \text{Original temperature}$$

When two or more objects at different temperatures are brought together in an isolated environment, they eventually reach the same temperature by the process of heat exchange. That is, warmer materials transfer heat to colder materials until their temperatures are the same.

$$- Q _ {x} = Q _ {\text{water}}$$ $$Q _ {x} = c _ {x} m _ {1} (27 - 82) = - 55 \cdot m _ {1} \cdot c _ {x}$$ $$Q _ {\text{water}} = c _ {H 2 O} m _ {H 2 O} (27 - 22) = 5 c _ {H 2 O} m _ {H 2 O}$$

where $c_{H2O} = 4186 \, \text{J/kg}{}^{\circ}\text{C}$, (specific heat capacity of water)

Thus,

$$55 c _ {x} m _ {1} = 5 c _ {H 2 O} m _ {H 2 O}$$

The specific heat of unknown substance is

$$c _ {x} = \frac {5 c _ {H 2 O} m _ {H 2 O}}{55 m _ {1}} = \frac {5 \cdot (4186 \, \text{J/kg}{}^{\circ}\text{C}) \cdot (0.025 \, \text{kg})}{55 \cdot (0.0025 \, \text{kg})} \approx 3805 \, \text{J/kg}{}^{\circ}\text{C}$$

No solid has such value of specific heat.

You need to clarify the input data, in particular the mass of block and water!

I guess that the value should be $380 \, \text{J/kg}{}^{\circ}\text{C}$

Answer: Brass

Answer provided by AssignmentExpert.com