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a volume of liquid weighing 25.0g is first heated in a flame. it is then poured into an insulated container with 350 ml of water at 45 oc (assume waters density is 1 g ml-1). when thermal equilibrium is reached (liquid and water at the same temperature), the temperature is 65 oc. note: the insulated container does not participate in the heat transfer process. cs for water is 4.18 j g-1 oc-1 . cs for the liquid is of 5.5 j g-1 oc-1 what is the temperature of the liquid when it is added to the water in oc?

User Vashum
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To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. The energy lost by the hot liquid is gained by the cold water, so we can write:

Q_lost = Q_gained

where Q is the amount of heat transferred. We can express Q as:

Q = m * Cs * ΔT

where m is the mass of the substance, Cs is its specific heat capacity, and ΔT is the change in temperature.

Let's start by calculating the amount of heat lost by the hot liquid. We know that it weighs 25.0 g, and we don't know its initial temperature, which we'll call T_l. We can express Q_lost as:

Q_lost = 25.0 g * Cs_l * (65 oC - T_l)

where Cs_l is the specific heat capacity of the liquid.

Next, let's calculate the amount of heat gained by the cold water. We know that it has a mass of 350 g and an initial temperature of 45 oC. We can express Q_gained as:

Q_gained = 350 g * Cs_w * (65 oC - 45 oC)

where Cs_w is the specific heat capacity of water.

Now we can set Q_lost equal to Q_gained and solve for T_l:

25.0 g * Cs_l * (65 oC - T_l) = 350 g * Cs_w * (65 oC - 45 oC)

Dividing both sides by 25.0 g * Cs_l, we get:

65 oC - T_l = 14.545

Subtracting 14.545 from 65 oC, we get:

T_l = 50.455 oC

Therefore, the initial temperature of the liquid was 50.455 oC when it was added to the water.
User Tomas Trdla
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