# A 17.5 g sample of metal at 125.0°C is placed in a calorimeter with 15.0 g of water at 25.0°C. if the temperature of the water rises to 30.0°C, what is the specific heat of the metal?

Jul 22, 2016

The specific heat of the meta is ${C}_{p} = 0.189 \frac{J}{{g}^{o} C}$

#### Explanation:

This is a thermo-equillibrium situation. We can use the equation

Loss of Heat of the Metal = Gain of Heat by the Water

$- {Q}_{m} = + {Q}_{w}$

$Q = m \Delta T {C}_{p}$

$Q =$ Heat
$m =$ mass
$\Delta T = \left({T}_{f} - {T}_{i}\right)$
${T}_{f} =$ Final Temp
${T}_{i} =$ Initial Temp
${C}_{P} =$ Specific Heat

Metal
Water
$m = 17.5 g$
${T}_{f} = {30.0}^{o} C$
${T}_{i} = {125.0}^{o} C$
${C}_{P} = x$

Water
$m = 15.0 g$
${T}_{f} = {30.0}^{o} C$
${T}_{i} = {25.0}^{o} C$
${C}_{P} = 4.184 \frac{J}{{g}^{o} C}$

$- {Q}_{m} = + {Q}_{w}$
$- \left[m \left({T}_{f} - T - i\right) {C}_{p}\right] = m \left({T}_{f} - T - i\right) {C}_{p}$

$- \left[17.5 g \left({30}^{o} C - {125}^{o} C\right) x\right] = 15 g \left({30}^{o} C - {25}^{o} C\right) 4.184 \frac{J}{{g}^{o} C}$

-[17.5g(-95^oC)x] = 15cancel(g)(5^ocancelC))4.184J/(cancel(g^oC))#

$\left(1662.5 {g}^{o} C\right) x = 313.8 J$

$\cancel{1662.5 {g}^{o} C} \frac{x}{\cancel{1662.5 {g}^{o} C}} = \frac{313.8 J}{1662.5 {g}^{o} C}$

${C}_{p} = 0.189 \frac{J}{{g}^{o} C}$