Ejercicio 2

Un termómetro de \( m = 55 \, \text{gr} \) y \( c = 0,2 \, \frac{\text{cal}}{\text{gr} \cdot \text{°C}} \) marca 15 °C. En ese punto se introduce en 300 g de agua y alcanza la temperatura final de la misma. Si el termómetro marca 44,4 °C, calcular la temperatura inicial del agua antes de introducir el termómetro. Despreciar las pérdidas de calor.
06- Física II Práctico TP2 - Parte 1 Calor 2021

1) Identificar los calores que intervienen:
Tenemos en cuenta el principio básico de calorimetría:
La sumatoria de todos los calores es igual a cero \[ \sum Q = 0 \] En este caso: \[ Q_{H2O} + Q_{term} = 0 \]
2) Desarrollar ecuación
Formulas \[ Q = m \cdot c \cdot \Delta T \] \[ m_{H2O} \cdot c_{H2O} \cdot \Delta T_{H2O} + m_{term} \cdot c_{term} \cdot \Delta T_{term} = 0 \] \[ m_{H2O} \cdot c_{H2O} \cdot (T_{1_{H2O}} - \color{orange} T_{0_{H2O}} \color{black}) + m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) = 0 \] Nuestra incógnita es \( \color{orange} T_{0_{H2O}} \color{black} \) por lo tanto despejamos: \[ m_{H2O} \cdot c_{H2O} \cdot (T_{1_{H2O}} - \color{orange} T_{0_{H2O}} \color{black}) = - m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) \] \[ T_{1_{H2O}} - \color{orange} T_{0_{H2O}} \color{black} = \frac {- m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) } {m_{H2O} \cdot c_{H2O}} \] \[ - \color{orange} T_{0_{H2O}} \color{black} = \frac {- m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) } {m_{H2O} \cdot c_{H2O}} - T_{1_{H2O}} \] Realizamos un artificio matemático : \[ - \color{orange} T_{0_{H2O}} \color{black} = \frac {- m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) } {m_{H2O} \cdot c_{H2O}} - \frac { T_{1_{H2O}} \cdot \color{purple} m_{H2O} \cdot c_{H2O} \color{black}} {\color{purple} m_{H2O} \cdot c_{H2O} \color{black}} \] \[ - \color{orange} T_{0_{H2O}} \color{black} = \frac { - m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) - T_{1_{H2O}} m_{H2O} \cdot c_{H2O} } {m_{H2O} \cdot c_{H2O}} \] \[ \color{orange} T_{0_{H2O}} \color{black} = \frac { m_{term} \cdot c_{term} \cdot (T_{1_{term}} - T_{0_{term}}) + T_{1_{H2O}} m_{H2O} \cdot c_{H2O} } {m_{H2O} \cdot c_{H2O}} \]

3) Reemplazar por los valores del ejercicio
El calor específico del agua es \( c_{H2O} = 1 \color{gray} [\frac{cal}{g \cdot °C}] \color{black} \)
Ver coeficientes \[ T_{0_{H2O}} = \frac { 55 \color{gray} [g] \color{black} \cdot 0,2 \color{gray} [\frac{cal}{g \cdot °C}] \color{black} \cdot (44,4 \color{gray} [°C] \color{black} - 15 \color{gray} [°C] \color{black}) + 300 \color{gray} [g] \color{black} \cdot 1 \color{gray} [\frac{cal}{g \cdot °C}] \color{black} \cdot 44,4 \color{gray} [°C] \color{black} } {300 \color{gray} [g] \color{black} \cdot 1 \color{gray} [\frac{cal}{g \cdot °C}] \color{black}} \] \[ T_{0_{H2O}} = \frac { 11 \color{gray} [g\frac{cal}{g \cdot °C}] \color{black} \cdot 29,4 \color{gray} [°C] \color{black} + 13.320 \color{gray} [g\frac{cal}{g \cdot °C}°C] } {300 \color{gray} [g\frac{cal}{g \cdot °C}] \color{black} } \] \[ T_{0_{H2O}} = \frac { 323,4 \color{gray} [g\frac{cal}{g \cdot °C}°C] \color{black} + 13.320 \color{gray} [g\frac{cal}{g \cdot °C}°C] } {300 \color{gray} [g\frac{cal}{g \cdot °C}] \color{black} } \] \[ T_{0_{H2O}} = \frac { 323,4 \color{gray} [cal] \color{black} + 13.320 \color{gray} [cal] \color{black} } {300 \color{gray} [\frac{cal}{°C}] \color{black} } \] \[ T_{0_{H2O}} = \frac {13.643,4 \color{gray} [cal] \color{black}} {300 \color{gray} [\frac{cal}{°C}] \color{black}} \] \[ T_{0_{H2O}} = 45,478 \color{gray} [cal\frac{°C}{cal}] \color{black} \] \[ \color{limegreen} \boxed{ \color{black} T_{0_{H2O}} = 45,478 \color{gray} [°C] \color{black} } \]