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Calculate the change in temperature of a gold ball (mass = 4.0 x 10⁻³ kg) when 53.88 J of heat is added. The specific heat capacity of gold is 1.29 x 102 J/kg·K.

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Answer

104.42 K\text{104.42 K}

Step-by-step solution

3 steps

Here's how to complete the problem:

  1. Identify the given values from your work.

    Heat added, Q=53.88 JQ = 53.88 \text{ J}
    Mass of gold ball, m=4.0×10−3 kgm = 4.0 \times 10^{-3} \text{ kg}
    Specific heat capacity of gold, cgold=1.29×102 J/kg⋅Kc_{gold} = 1.29 \times 10^2 \text{ J/kg}\cdot\text{K}

  2. State the formula for heat transfer.

    The formula for heat transfer is:

    Q=mcΔTQ = mc\Delta T

    where QQ is the heat added, mm is the mass, cc is the specific heat capacity, and ΔT\Delta T is the change in temperature.

  3. Substitute the known values into the formula and solve for ΔT\Delta T.

    53.88 J=(4.0×10−3 kg)⋅(1.29×102 J/kg⋅K)⋅ΔT53.88 \text{ J} = (4.0 \times 10^{-3} \text{ kg}) \cdot (1.29 \times 10^2 \text{ J/kg}\cdot\text{K}) \cdot \Delta T

    First, calculate the product of mass and specific heat capacity:

    (4.0×10−3)⋅(1.29×102)=4.0⋅1.29⋅10−3+2=5.16⋅10−1=0.516 J/K(4.0 \times 10^{-3}) \cdot (1.29 \times 10^2) = 4.0 \cdot 1.29 \cdot 10^{-3+2} = 5.16 \cdot 10^{-1} = 0.516 \text{ J/K}

    Now, substitute this back into the equation:

    53.88 J=(0.516 J/K)⋅ΔT53.88 \text{ J} = (0.516 \text{ J/K}) \cdot \Delta T ΔT=53.88 J0.516 J/K\Delta T = \frac{53.88 \text{ J}}{0.516 \text{ J/K}} ΔT=104.4186 K\Delta T = 104.4186 \text{ K}

    The change in temperature is 104.42 K\text{104.42 K}.

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