Question

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g⋅K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.3°C .
(a)- Determine the amount of heat, in J, lost by the copper block. Please enter the absolute amount of heat lost by Cu, without a minus sign.
(b)- Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g⋅K.
(c)-The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam → cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat
capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K.
Expressyour answer using two significant figures.
(d)-What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

Answer 1

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C  

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J \cdot K ^{-1} g ^{-1} }\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J \cdot K ^{-1} g ^{-1} }\times \Delta T_{\text{w}}\\\text{47.7 J \cdot K ^{-1} }\times \Delta T_{\text{Cu}}& = &-\text{585 J \cdot K ^{-1} g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.