A 35.41 g piece of copper at 97°C was placed in a sample of water at 21.65°C. The temperature of the water increased to 22.87°C. What mass of water was the copper added to?

197.98g

Explanation:

This question can be calculated using the formula: Q = mcΔT

Where Q= heat transfer

M= mass

C= specific capacity

ΔT= change in temperature

Heat from the copper metal is lost to the water when placed in it, hence, we have:

Q(water) = -Q(copper)

The final temperature of the water is equally to the final temperature of the copper= 22.87°C.

Hence, mcΔT (water) = (-) mcΔT (copper)

Note that, specific heat (C) of water and copper are 4.184 J/g°C and 0.385 J/g°C respectively.

Therefore;

= (m) (4.184) (22.87-21.65) = - {(35.41) (0.385) (22.87-97)}

= m × 4.184 × 1.22 = - {35.41 × 0.385 × -74.13)}

5.10448m = - {-1010.6031}

5.10448m = 1010.6031

m = 1010.6031 ÷ 5.10448

m = 197.98

Therefore, the mass of water copper was added to is 197.98g

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explanation:

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