(b) Some pieces of stainless steel cubes at -18 °C are added into 250 g of water at 15 °C. The mass of each steel cube is 50 g. The specific heat capacities of stainless steel and water are 502 J kg⁻¹ °C⁻¹ and 4200 J kg⁻¹ °C⁻¹ respectively. If the final temperature of the water is 9.7 °C, how many cubes are added?
Answer (1)
To find out how many stainless steel cubes were added to the water, we'll use the principle of conservation of energy, assuming no heat is lost to the surroundings. The heat lost by water will be equal to the heat gained by the stainless steel cubes.
Step-by-Step Calculation:
Calculate the Heat Loss by Water:
The formula for heat transfer is:
Q = m c Δ T
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity
Δ T is the change in temperature
For the water:
Mass m = 250 g = 0.250 kg
Specific heat capacity c = 4200 J kg − 1 ° C − 1
Initial temperature T i = 15° C
Final temperature T f = 9.7° C
Change in temperature Δ T = T f − T i = 9.7 − 15 = − 5.3° C
So, the heat lost by water is:
Q water = 0.250 × 4200 × ( − 5.3 )
Q water = − 5565 J
Calculate the Heat Gain by Stainless Steel Cubes:
For each cube of stainless steel:
Mass m = 50 g = 0.050 kg
Specific heat capacity c = 502 J kg − 1 ° C − 1
Initial temperature T i = − 18° C
Final temperature T f = 9.7° C
Change in temperature Δ T = T f − T i = 9.7 − ( − 18 ) = 27.7° C
So, the heat gained by one cube is:
Q steel = 0.050 × 502 × 27.7
Q steel = 695.77 J
Determine the Number of Cubes:
Let's assume n is the number of cubes. The total heat gained by all cubes will be equal to the total heat lost by the water:
n × 695.77 = 5565
Solving for n :
n = 695.77 5565 ≈ 8
Therefore, 8 stainless steel cubes are added to the water to result in the final temperature of 9.7 ° C .
