Questions: A 5.00 kg block of metal with c=650 J /(kgC) at 80.0°C comes in contact with a 1.25 kg glass block at 20.0°C. They come to equilibrium at 63.9°C. What is the specific heat of the glass? (Unit =J /(kgC))

A 5.00 kg block of metal with c=650 J /(kg*C) at 80.0°C comes in contact with a 1.25 kg glass block at 20.0°C. They come to equilibrium at 63.9°C. What is the specific heat of the glass?

(Unit =J /(kg*C))

Transcript text: A 5.00 kg block of metal with $\mathrm{c}=650 \mathrm{~J} /\left(\mathrm{kg}^{*} \mathrm{C}\right)$ at $80.0^{\circ} \mathrm{C}$ comes in contact with a 1.25 kg glass block at $20.0^{\circ} \mathrm{C}$. They come to equilibrium at $63.9^{\circ} \mathrm{C}$. What is the specific heat of the glass? \[ \left(\text { Unit }=\mathrm{J} /\left(\mathrm{kg}^{*} \mathrm{C}\right)\right) \]

Solution

Solution Steps

Step 1: Define the Problem and Known Variables

We need to find the specific heat capacity of the glass block. The known variables are:

Mass of metal block, $ m_1 = 5.00 \, \text{kg} $
Specific heat capacity of metal, $ c_1 = 650 \, \text{J/(kg} \cdot \text{°C)} $
Initial temperature of metal, $ T_{1i} = 80.0 \, \text{°C} $
Mass of glass block, $ m_2 = 1.25 \, \text{kg} $
Initial temperature of glass, $ T_{2i} = 20.0 \, \text{°C} $
Final equilibrium temperature, $ T_f = 63.9 \, \text{°C} $

Step 2: Apply the Principle of Conservation of Energy

The heat lost by the metal block will be equal to the heat gained by the glass block. This can be expressed as: \[ m_1 c_1 (T_{1i} - T_f) = m_2 c_2 (T_f - T_{2i}) \] where $ c_2 $ is the specific heat capacity of the glass block that we need to find.

Step 3: Substitute Known Values into the Equation

Substitute the known values into the equation: \[ 5.00 \times 650 \times (80.0 - 63.9) = 1.25 \times c_2 \times (63.9 - 20.0) \]

Step 4: Simplify and Solve for $ c_2 $

First, calculate the temperature differences: \[ 80.0 - 63.9 = 16.1 \, \text{°C} \] \[ 63.9 - 20.0 = 43.9 \, \text{°C} \]

Now, substitute these values back into the equation: \[ 5.00 \times 650 \times 16.1 = 1.25 \times c_2 \times 43.9 \]

Calculate the left-hand side: \[ 5.00 \times 650 \times 16.1 = 52325 \, \text{J} \]

Now solve for $ c_2 $: \[ 52325 = 1.25 \times c_2 \times 43.9 \] \[ c_2 = \frac{52325}{1.25 \times 43.9} \] \[ c_2 = \frac{52325}{54.875} \] \[ c_2 \approx 953.4 \, \text{J/(kg} \cdot \text{°C)} \]