# Question e3133

Jul 3, 2017

The specific heat is: $\left(0.812 \text{J")/("g of Metal} \cdot {C}^{\circ}\right)$

#### Explanation:

We are told that $100.0 \text{g of H"_2"O}$ went from ${20}^{\circ} C$ to ${25}^{\circ} C$. The reference Specific Heat gives us the equation for how much heat energy was absorbed by the water:

$Q = c m \Delta T \text{ [1]}$

where c= "the specific heat of water" = (4.139"J")/("g H"_2"O"*"C"^@)#, $m = \text{the mass" = 100.0"g of H"_2"O}$, and $\Delta T = \text{the change in temperature} = {25}^{\circ} C - {20}^{\circ} C$

Substitute these values into equation [1]:

$Q = \left(4.139 \text{J")/("g H"_2"O"*"C"^@)(100.0"g H"_2"O}\right) \left({25}^{\circ} C - {20}^{\circ} C\right)$

The water absorbed:

$Q = 2069.5 \text{J}$

of heat energy.

The energy came from the metal and it obeys the same equation, equation [1]:

$Q = c m \Delta T \text{ [1]}$

We know everything in this equation, except the value of, c -- the specific heat for the metal, therefore, we can substitute in the known values and then solve for c:

We know that $Q = - 2069.5 \text{J}$, because the energy that went into the water came out from the hot metal:

$- 2069.5 \text{J"= cmDeltaT" [1.1]}$

We know that $m = 30 \text{g of Metal}$

$- 2069.5 \text{J"= c(30"g of Metal")DeltaT" [1.2]}$

We know that the temperature of the metal went from ${110}^{\circ} \text{C}$ to the same temperature as the water, ${25}^{\circ} \text{C}$, therefore, $\Delta T = {25}^{\circ} \text{C"-110^@"C}$:

$- 2069.5 \text{J"= c(30"g of Metal")(25^@"C"-110^@"C")" [1.3]}$

To solve for c, we divide by the mass and the change in temperature:

$c = \left(- 2069.5 \text{J")/((30"g of Metal")(25^@"C"-110^@"C")}\right)$

$c = \left(0.812 \text{J")/("g of Metal} \cdot {C}^{\circ}\right)$

This is the specific heat of the metal.