MERIC RESPONSE QUESTIONS 29-30=Total/4points ne the temperature change of 300 grams of water when 1000 joules of energy is ad __ __ __ C_(water)=4.19J/g^circ C rd your three-digit answer on the numerical-response answer sheet provided.)

Here's how to solve this problem:<br /><br />1. **Recall the formula that relates heat energy (Q), mass (m), specific heat capacity (c), and temperature change (ΔT):**<br /><br /> Q = mcΔT<br /><br />2. **Rearrange the formula to solve for the temperature change (ΔT):**<br /><br /> ΔT = Q / (mc)<br /><br />3. **Plug in the given values:**<br /><br /> * Q = 1000 J<br /> * m = 300 g<br /> * c = 4.19 J/g°C<br /><br /> ΔT = 1000 J / (300 g * 4.19 J/g°C)<br /><br />4. **Calculate the temperature change:**<br /><br /> ΔT ≈ 0.796 °C<br /><br />5. **Round to the nearest hundredth:**<br /><br /> ΔT ≈ 0.80 °C<br /><br />Therefore, the temperature change of the water is approximately 0.80°C. Since the question asks for a three-digit answer, you would enter *0.80* or *080* on the numerical response sheet.<br />